ITSA E-tutor Code Exchange

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貼 CODE 請按照下方提供的固定格式（按左上角「筆」或「視窗分割」按鈕就可以看到）

首先找到程式練習的題庫，沒有的話用 h2 （兩個#）開新的區塊。
使用 h3 （三個#）貼上題名(再放置連結)，E-tutor格式如下。
使用下面這個範本來貼程式碼（Java可以替換成其他 HackMD 支援的語言，如 C++ 為 CPP ，"="是提供行數標示，如果同一題有不同語言的解法，應放在同一題的區塊下，用 h4 （四個#）標示使用何種語言，同語言不同解法可用 h5 （五個#）標示解法一、解法二⋯⋯等等。
可標示程式碼提供者。
本頁面使用 C 寫的 Code 在格式部分請選擇「cpp=」。
右側有選單 (雙欄編輯模式時會收至右下角) 可以快速跳至欲查看的區塊，不用慢慢滑。
程式碼力求可讀性高(因為看到自己的黑歷史)。
善用註解，另可額外補充解決此題應知資訊。
看看別人、看看自己的程式碼，不妨想想 TQM (Total Quality Control) 「完全品質控制」的概念，即「重來一次，怎麼做會更好？」一次次的思考，伴隨著一次次的提升。
如果有能力的話，請嘗試不同的解法；也別被別人的想法拘束住。


PASTE YOUR CODE HERE

我的編譯環境說明：
C++ 版本使用 C++ 14，編譯器為 clang/llvm
OS: macOS Mojave 10.14.2
Howard Guo

每天都要記得 Coding，C、C++、Java至少要碰其中一個!
Sat, Jul 20, 2019 1:16 AM
風思

Mathematics I

[C_MM01-易] 計算梯型面積

C Version

Howard Guo














#include <stdio.h>
#include <stdlib.h>

/* run this program using the console pauser or add your own getch, system("pause") or input loop */

int main(int argc, char *argv[]) {
	int a;
	int b;
	int c;
	scanf("%d %d %d", &a, &b, &c );
	double result = (a+b)*c/2.0;
	printf("Trapezoid area:%.1f\n", + result );
	return 0;
}

風思









#include <stdio.h>
#include <stdlib.h>

int main(int argc, char *argv[]) {
	int a , b , c;
	scanf("%d %d %d",&a,&b,&c);
	printf("Trapezoid area:%.1f\n",(a+b)*c/2.0);
	return 0;
}

Java Version

AfreecaKing











import java.util.*;

class Main {
  public static void main(String[] args) {
    Scanner sc =new Scanner(System.in);
    int up=sc.nextInt();
    int down=sc.nextInt();
    int high=sc.nextInt();
    System.out.println("Trapezoid area:"+((up+down)*high/2.0));  
  }
}

[C_MM02-易] 計算三角形面積

Java Version

AfreecaKing










import java.util.*;

class Main {
  public static void main(String[] args) {
    Scanner sc =new Scanner(System.in);
    int under=sc.nextInt();
    int high=sc.nextInt();
    System.out.println(under*high/2.0);
  }
}

[C_MM03-易] 兩數總和

Java Version

AfreecaKing












import java.util.*;

class Main {
  public static void main(String[] args) {
    Scanner sc =new Scanner(System.in);
    while(sc.hasNext()){              
      int n=sc.nextInt();
      int m=sc.nextInt();
      System.out.println(n+m);
    }
  }
}

[C_MM04-易] 計算總和、乘積、差、商和餘數

Java Version

AfreecaKing
注意餘數不能是負的




















import java.util.*;

public class Main {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner sc =new Scanner(System.in); 
	    int a=sc.nextInt(); 
	    int b=sc.nextInt(); 
	    System.out.printf("%d+%d=%d\n",a,b,a+b); 
	    System.out.printf("%d*%d=%d\n",a,b,a*b); 
	    System.out.printf("%d-%d=%d\n",a,b,a-b);
	    if(a%b>=0)
	    	System.out.printf("%d/%d=%d...%d\n",a,b,a/b,a%b); 
	    else
	    	System.out.printf("%d/%d=%d...%d\n",a,b,a/b-1,a%b+b);

	}

}

[C_MM05-易] 計算正方形面積

C Version

風思
當初卡很久在四捨五入的問題




















#include <stdio.h> 
#include <stdlib.h> 
#include <math.h> 
int main()
{
    double a,b,c;
    
    scanf("%lf",&a);
    b=pow(a,2); //平方
    //printf("%.1f\n",b); 
    c= (int)(b); //取整數,floor(b,2);
    //printf("%.0f\n",c); 
    if(b-c>=0.05) printf("%.1f",b+0.01); /*取小數第二位,若大於0.05,則加上0.01,
    又因為c語(可)言(誤)本身會五捨六入,這樣恰好可以四捨五入,這花了我好多時間...*/
    else printf("%.1f",b);
    
    printf("\n");

    return 0;
}

Java Version

解法一

風思
相同的格式化輸出，java四捨五入；C五捨六入













import java.util.*;

public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		while (sc.hasNext()) {
			double n = sc.nextDouble();
			System.out.printf("%.1f\n", Math.pow(n, 2));
		}
		sc.close();
	}
}

解法二

風思
當初寫C要是知道這招，就好啦!













import java.util.*;

public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		while(sc.hasNext()) {
			double n=sc.nextDouble();
			System.out.println(Math.round(n*n*10)/10.0);
		}
		sc.close();
	}
}

[C_MM06-易] 英哩轉公里

Java Version

AfreecaKing








import java.util.*;
public class Main {
  public static void main(String[] args) {
    Scanner sc=new Scanner(System.in);
    double n=sc.nextInt();
    System.out.println(Math.round(1.6*n*10.0)/10.0);
  }
}

[C_MM07-易] 計算平方值與立方值

Java Version

AfreecaKing








import java.util.*;
class Main {
  public static void main(String[] args) {
    Scanner sc=new Scanner(System.in);
    int n=sc.nextInt();
    System.out.printf("%d %d %d\n",n,n*n,n*n*n);
  }
}

[C_MM08-易] 計算兩數和的平方值

Java Version

Afreecaking









import java.util.*;
class Main {
  public static void main(String[] args) {
    Scanner sc=new Scanner(System.in);
    int n=sc.nextInt();
    int m=sc.nextInt();
    System.out.println((n+m)*(n+m));
  }
}

[C_MM09-易] 計算 i 次方的值

Java Version

AfreecaKing

















import java.util.*;
class Main {
  public static void main(String[] args) {
    Scanner sc=new Scanner(System.in);
    int i=sc.nextInt();
    int sum=1;
    if(i>31)
      System.out.println("Value of more than 31");
    else
      {
        for(int j=1;j<=i;j++)
          sum=sum*2;
        System.out.println(sum);
      }
   
  }
}

[C_MM10-易] 攝氏溫度轉華式溫度

Java Version

AfreecaKing
強烈建議不要用C寫，浮點數會有誤差












import java.util.*;

public class Main {

	public static void main(String[] args) {
		Scanner sc =new Scanner(System.in);
		double n=sc.nextDouble();
		System.out.println(Math.round((n*9/5+32)*10.0)/10.0);
		sc.close();
	}

}

C Version

AfreecaKing
用round來處理浮點數誤差，很久以前寫的可讀性不高














#include <stdio.h>  
#include <stdlib.h>  
#include <cmath>
  
int main()  
{  
    double n;
    scanf("%lf",&n);
    double X,f;
    f=n*(9/5.0)+32;
    X=round(10*f)/10.0;
    printf("%.1lf\n",X);
    return 0;
}

[C_MM11-易] 購票計算

C Version

風思
天罰版，可讀性幾乎為0






































#include <stdio.h>

int main ()
{
  int x, t = 0, m = 0, f = 0, o = 0;
 //x=初始值,t=NT10數量,m=過渡運算值,f=NT5數量,o=NT1數量;
  scanf ("%d", &x);
  if (x >= 10)
    {				
      t = x / 10;
      m = x - 10 * t;

      if (m >= 5)
	{
	  f = m / 5;
	  o = m - 5 * f;

	}
      else
	{
	  o = m;
	}
    }
  if (x >= 5 && x < 10)
    {
      f = x / 5;
      o = x - 5;

    }
  if (x < 5)
    {
      o = x;
    }
  printf ("NT10=%d\n", t);
  printf ("NT5=%d\n", f);
  printf ("NT1=%d\n", o);
  return 0;
}

Java Version

風思
好多了


















import java.util.*;
public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		while(sc.hasNext()) {
			int n = sc.nextInt();
			int n10= n/10;
			int n5= (n%10)/5;
			int n1= n%5;
			System.out.println("NT10="+n10);
			System.out.println("NT5="+n5);
			System.out.println("NT1="+n1);
		}
        sc.close();
	}

}

Java Version

AfreecaKing


















import java.util.*;

public class Main {

	public static void main(String[] args) {
		Scanner sc =new Scanner(System.in);
		int n=sc.nextInt();
		int a,b,c; //寫C的習慣了
		a=n/10;
		b=n%10/5;
		c=n%5;
		System.out.println("NT10="+a);
		System.out.println("NT5="+b);
		System.out.println("NT1="+c);
		sc.close();
	}

}

[C_MM12-易] 相遇時間計算

Java Version

AfreecaKing












import java.util.*;

public class Main {

	public static void main(String[] args) {
		Scanner sc =new Scanner(System.in);
		int n=sc.nextInt();
		System.out.println((int)Math.ceil(n/0.238));
		sc.close();
	}

}

[C_MM13-易] 停車費計算

Java Version

AfreecaKing
將分鐘換成秒來計算



















import java.util.*;

public class Main {
  public static void main(String[] args) {
   Scanner sc =new Scanner(System.in);
   int n1=sc.nextInt();
   int m1=sc.nextInt();
   int n2=sc.nextInt();
   int m2=sc.nextInt();
   sc.close();
   int sum=(n2*60+m2)-(n1*60+m1);
   if(sum<=120)
    System.out.println(sum/30*30);
   else if(sum>120 && sum<240)
    System.out.println(120+((sum-120)/30*40));
   else
    System.out.println(280+((sum-240)/30*60));
  }
}

[C_MM14-易] 計算時間的組合

Java Version

AfreecaKing
















import java.util.*;

public class Main {
  public static void main(String[] args) {
   Scanner sc =new Scanner(System.in);
   int second=sc.nextInt();
   sc.close();
   System.out.println(+second/86400+" days");
   System.out.println(+second%86400/3600+" hours");
   System.out.println(+second%3600/60+" minutes");
   System.out.println(+second%60+" seconds");
  }
}
//一天86400秒
//一小時3600秒
//一分鐘60秒

[C_MM15-易] 判斷座標是否在正方形的範圍內

Java Version

AfreecaKing














import java.util.*;
public class Main {

	public static void main(String[] args) {
		Scanner sc =new Scanner(System.in);
		int x=sc.nextInt();
		int y=sc.nextInt();
		sc.close();
		if(x<=100 && y<=100)
			System.out.println("inside");
		else
			System.out.println("outside");
	}
}

[C_MM16-易] 判斷座標是否在圓形的範圍內

Java Version

AfreecaKing















import java.util.*;
public class Main {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner sc =new Scanner(System.in);
		int x=sc.nextInt();
		int y=sc.nextInt();
		if(x<=100 && y<=100 && x*x+y*y<=10000)
			System.out.println("inside");
		else
			System.out.println("outside");
	}

}

[C_MM17-易] 求最大公因數

Java Version

AfreecaKing
也可以用輾轉相除法，請上網路自行搜尋
因為不好解釋，要用到遞迴的概念






















import java.util.*;
class Main {
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    int n=sc.nextInt();
    int m=sc.nextInt();
    if(n>m)
      for(int i=m;i>0;i--)
        if(n%i==0 && m%i==0)
        {
          System.out.println(i);
          break;
        }
    if(m>n)
      for(int i=n;i>0;i--)
        if(n%i==0 && m%i==0)
        {
          System.out.println(i);
          break;
        }
  }
}

[C_MM18-易] 十進制轉二進制

Java Version

風思
~n為n的ones' complement（一的補數）

































import java.util.*;
//[C_MM18-易] 十進制轉二進制
public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int[] result = new int[8];
		if (n < 0) {
			int negn = ~n; //負數變成其一補數，方便處理
			for (int i = 0; i < 8; i++) {
				result[7 - i] = negn % 2;
				negn /= 2;
			}
			for (int i = 0; i < 8; i++) {
				if (result[i] == 0)
					result[i] = 1;
				else
					result[i] = 0;
			}
		} else {
			for (int i = 0; i < 8; i++) {
				result[7 - i] = n % 2;
				n /= 2;
			}
		}
		for (int i = 0; i < 8; i++)
			System.out.print(result[i]);
		System.out.println();
		sc.close();

	}
}

[C_MM19-易] 電話費計算

Java Version

AfreecaKing















import java.util.*;

class Main {
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    int time=sc.nextInt();
    sc.close();
    if(time<=800)
      System.out.println(time*0.9);
    else if(time>800 && time<1500)
      System.out.println(Math.round(time*0.9*0.9*10.0)/10.0);
    else
      System.out.println(Math.round(time*0.9*0.79*10.0)/10.0);
  }
}

[C_MM36-易] 季節判定

C++ Version

這題也是用笨笨 switch 硬幹， There should be a better solution even it's a simple problem.
Howard Guo
















































#include<iostream>
using namespace std;
int main(){
    int month;
    cin >> month;

    switch (month)
    {
        case 3:
            cout << "Spring" << endl;
            break;
        case 4:
            cout << "Spring" << endl;
            break;
        case 5:
            cout << "Spring" << endl;
            break;
        case 6:
            cout << "Summer" << endl;
            break;
        case 7: 
            cout << "Summer" << endl;
            break;
        case 8:
            cout << "Summer" << endl;
            break;
        case 9:
            cout << "Autumn" << endl;
            break;
        case 10:
            cout << "Autumn" << endl;
            break;
        case 11:
            cout << "Autumn" << endl;
            break;
        case 12:
            cout << "Winter" << endl;
            break;
        case 1:
            cout << "Winter" << endl;
            break;
        case 2:
            cout << "Winter" << endl;
            break;
    }
    return 0;
}

C Version

風思
嘿嘿嘿，比↑少了粉多break跟輸出~
































#include <stdio.h>
#include <stdlib.h>

int main(int argc, char *argv[]) {
	int n;  
      
    scanf("%d",&n);  
      
    switch (n){  
        case 3:  
        case 4:  
        case 5:  
           printf("Spring\n");  
           break;  
        case 6:  
        case 7:  
        case 8:  
          printf("Summer\n");  
          break;  
        case 9:  
        case 10:  
        case 11:  
          printf("Autumn\n");         
          break;  
        case 12:  
        case 1:  
        case 2:  
            printf("Winter\n");       
          break;  
    }
	return 0;
}

Java Version

風思



































import java.util.*;
public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		while(sc.hasNext()) {
			int n = sc.nextInt();
			switch (n){  
	        case 3:  
	        case 4:  
	        case 5:  
	           System.out.println("Spring");  
	           break;  
	        case 6:  
	        case 7:  
	        case 8:  
	        	System.out.println("Summer");  
	          break;  
	        case 9:  
	        case 10:  
	        case 11:  
	        	System.out.println("Autumn");         
	          break;  
	        case 12:  
	        case 1:  
	        case 2:  
	        	System.out.println("Winter");       
	          break;  
	    }
		    
		}
        sc.close();
	}

}

[C_MM37-易] 判斷座標位於何處

C++ Version

Howard Guo


















#include <iostream>
using namespace std;
int main(void){
	int x, y;
	cin >>x >> y;
	
	if(x==0 || y==0){
		if(x==0 && y==0) cout << "Original" << endl;
		else if(x==0 && y!=0) cout << "y-axis" << endl;
		else if(y==0 && x!=0) cout << "x-axis" << endl;
	}else if(x>0 && y >0)	cout << "1st Quadrant" << endl;
	else if(x<0 && y>0)		cout << "2nd Quadrant" << endl; 
	else if(x<0 && y<0)		cout << "3rd Quadrant" << endl;
	else if(x>0 && y<0)		cout << "4th Quadrant" << endl;
	
	
	return 0;
}

C Version

風思


























#include <stdio.h>

int main()
{
    int x,y;
    
    scanf("%d %d",&x,&y);
    
    if(x==0&&y==0) printf("Origin\n");
    
    if(x==0&&y!=0) printf("y-axis\n");
    
    if(x!=0&&y==0) printf("x-axis\n");
    
    if(x>0&&y>0) printf("1st Quadrant\n");
    
    if(x<0&&y>0) printf("2nd Quadrant\n");
    
    if(x<0&&y<0) printf("3rd Quadrant\n");
    
    if(x>0&&y<0) printf("4th Quadrant\n");
    
    

    return 0;
}

[C_MM38-易] 判斷3整數是否能構成三角形之三邊長

C++ Version

Howard Guo


























#include <iostream>
using namespace std;
int main(int argc, char const *argv[])
{
    int a, b, c;
    while (cin >> a >> b >> c)
    {
        if ((a + b) > c)
        {
            if ((b + c) > a)
            {
                if ((a + c) > b)
                {
                    cout << "fit" << endl;
                }
                else
                    cout << "unfit" << endl;
            }
            else
                cout << "unfit" << endl;
        }
        else
            cout << "unfit" << endl;
    }
    return 0;
}

C Version

風思















#include <stdio.h>

int main()
{
    int x,y,z;
    
    scanf("%d %d %d",&x,&y,&z);
    
    if(x+y>z&&x+z>y&&y+z>x){
        printf("fit\n");
    }else{
        printf("unfit\n");
    }
    return 0;
}

Java Version

風思






















import java.util.*;
public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		while(sc.hasNext()) {
			int x = sc.nextInt();
			int y = sc.nextInt();
			int z = sc.nextInt();
		    
		    if(x+y>z&&x+z>y&&y+z>x){
		        System.out.println("fit");
		    }else{
		    	System.out.println("unfit");
		    }
		    
		}
        sc.close();
	}

}

[C_MM39-易] 判斷是何種三角形

C++ Version

Howard Guo






















































#include <iostream>
using namespace std;
void tri(int, int, int);
int main(int argc, char const *argv[])
{
    int a, b, c;
    while (cin >> a >> b >> c)
    {
        tri(a, b, c);
    }
    return 0;
}

void tri(int a, int b, int c)
{
    int temp;
    for (int i = 1; i <= 3; i++)
    {
        if (a > b)
        {
            temp = a;
            a = b;
            b = temp;
        }
        else if (b > c)
        {
            temp = b;
            b = c;
            c = temp;
        }
        else if (a > c)
        {
            temp = a;
            a = c;
            c = temp;
        }
    }
    if(a+b<=c){
        cout << "Not Triangle" << endl;
    }
    else if ((a * a + b * b) == c * c)
    {  //直角三角形
        cout << "Right Triangle" << endl;
    }
    else if ((a * a + b * b) < c * c)
    {  //鈍角三角形
        cout << "Obtuse Triangle" << endl;
    }
    else if ((a * a + b * b) > c * c)
    {  //銳角三角形
        cout << "Acute Triangle" << endl;
    }
}

C Version

風思
















#include <stdio.h>

int main()  
{  
   int a,b,c,a2,b2,c2;  
   scanf("%d %d %d",&a,&b,&c);  
   a2=a*a;  
   b2=b*b;  
   c2=c*c;  
   if(a+b>c && a+c>b && b+c>b) {  
   if(a2+b2==c2 || a2+c2==b2 || b2+c2==b2) printf("Right Triangle\n");  
   if(a2+b2<c2 || a2+c2<b2 || b2+c2<b2) printf("Obtuse Triangle\n");  
   if(a2+b2>c2 && a2+c2>b2 && b2+c2>b2) printf("Acute Triangle\n");  
   }else printf("Not Triangle\n");  
    return 0;  
}

[C_MM40-易] 1~N之間的總和

C++ Version

Howard Guo
















#include<iostream>
using namespace std;
int main(void){
    int n, sum=0;
    cin >> n;
    for(int i=1;i<=n;i++){
        sum += i;
    }

    for(int i=1;i<=n;i++){
        cout << i;
        if(i != n) cout << " + ";
    }
    cout << " = " << sum << endl;
    return 0;
}

C version

風思
















#include <stdio.h> 
 
int main() 
{ 
   int a,i,sum; 
   scanf("%d",&a); 
   for(i=1;i<=a;i++){ 
       sum+=i; 
      if(i==1) printf("%d ",i); 
      else printf("+ %d ",i); 
   } 
    
   printf("= %d",sum); 
   printf("\n");
    return 0; 
}

[C_MM44-易] The Numbers

C Version

風思
scanf("%[^\n]",&N)為可輸入空白之字串表示方法






















include <stdio.h>

int main()
{
    char N[11]; //欲將兩數合併為一字串,中有空格後有'\0',
    //所以=>N[2+1+7+1];
    char M[8]; //預備給第二數儲存=>用來與第一數比對;
    short i,times; //宣告times紀錄次數;
    scanf("%[^\n]",&N); //輸入兩數;
    //printf("%s",N);
    for(i=3;i<=9;i++){ //由於第二數由N[3]開始;
        M[i-3]=N[i]; //複製;
    }
    //printf("%s",M);
   for(i=0;i<=6;i++){ //兩兩比對,至多比對7-1=6次;
        if(M[i]==N[0] && M[i+1]==N[1]) times++;
   }
    printf("%d",times);
    printf("\n");
    //想好久...2019/1/30 10:10;
    return 0;
}

[C_MM48-易] F91

C++ Version

Howard Guo




































#include <iostream>
#include<cstring>  //To use the fuction "memset".
using namespace std;
int f91(int);
int main(int argc, char const *argv[])
{
    int k; //根據題意：有k個測試資料
    cin >> k;
    int arr[k];
    memset(arr, 0, k); //把陣列arr清0
    for (int i = 0; i < k; i++)
    {
        cin >> arr[i];
    }
    for (int i = 0; i < k; i++)
    {
        cout << f91(arr[i]) << endl;
    }
    return 0;
}

int f91(int n)
{
    if (n >= 101)
    {
        return n - 10;
    }
    else if (n <= 100)
    {
        return f91(f91(n + 11));
    }
    else
    {
        return 0;
    }
}

C Version

風思




























#include <stdio.h>
int f91(int n){
    if(n>=1 && n<=100000){
        if(n<=100) return f91(f91(n+11));
        if(n>=101) return n-10;
    }
    
}

int main()
{
    int k,i;
    scanf("%d",&k);
    int a[k];
    if(k>=1 && k<=10){
        for(i=0;i<=k-1;i++){
         if(i!=k-1) scanf("%d ",&a[i]);
         else scanf("%d",&a[i]);
        }
        for(i=0;i<=k-1;i++){
         if(i!=k-1) printf("%d\n",f91(a[i]));
         else printf("%d",f91(a[i]));
        }
    }
    //可惡的格式;
    printf("\n");
    return 0;
}

Arrayc I

[C_AR01-易] 一維陣列反轉 I

C++ Version

Howard Guo









































#include <iostream>
#include <cstring>
using namespace std;
int main(int argc, char const *argv[])
{
    int arr[100];
    int arr_count = 0;
    string input;
    while (getline(cin, input))
    {
        memset(arr, 0, 100); //陣列清0
        arr_count = 0;       //計數器歸零，不然第二次迴圈開始就會亂吐東西
        for (int i = 0; i < input.length(); i++)
        {
            if (input[i + 1] != ' ' && input[i] != ' ' && i != input.length() - 1)
            {
                //cout << input[i] << "a" << endl;                                              For purpose of debugging
                arr[arr_count] = arr[arr_count] * 10 + (int(input[i]) - '0') * 10;
            }
            else if (input[i + 1] == ' ' || i == input.length() - 1)
            {
                //cout << input[i] << "q" << endl;
                arr[arr_count] += int(input[i]) - '0';
                //cout << "value= " << arr[arr_count] << " @ index= " << arr_count << endl;   For purpose of debugging
                arr_count++;
                //cout << "new_count" << arr_count << endl;                                   For purpose of debugging
            }
        }

        //印
        for (int i = arr_count - 1; i >= 0; i--)
        {
            if (i == arr_count - 1)
                cout << arr[i];
            else
                cout << " " << arr[i];
        }
        cout << endl;
    }
    return 0;
}

Java Version

風思
未來你會看到很多spilt();




















import java.util.*;

public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		while (sc.hasNext()) {
			String st = sc.nextLine();
			String[] tokens = st.split(" ");
			for (int i = tokens.length - 1; i >= 0; i--) {
				if (i == tokens.length - 1) 
					System.out.print(tokens[i]);
				else System.out.print(" "+tokens[i]);

			}
            System.out.println();
		}
		sc.close();
	}
}

[C_AR02-易] 一維陣列反轉 II

C Version

風思



















#include <stdio.h>
#include <stdlib.h>

int main(int argc, char *argv[]) {
	int a[6];
    int i;
    for(i=0;i<6;i++){
    	scanf("%d",&a[i]);
	}
	for(i=5;i>=0;i--){
		if(i==5){
			printf("%d",a[i]);
		}else{
			printf(" %d",a[i]);
		}
	}
	printf("\n");
	return 0;
}

Howard Guo






















#include <stdio.h>
#include <stdlib.h>

/* run this program using the console pauser or add your own getch, system("pause") or input loop */

int main(int argc, char *argv[]) {
	int a[6];
	int i=0;
	for(i=0;i<6;i++){
		scanf("%d", &a[i]);
	}
	
	for(i=5;i>=0;i--){
		if(i==5){  //要注意不能用= 
			printf("%d", a[i]);
		}else{
			printf(" %d", a[i]);
		}
	}
	printf("\n");
	return 0;
}

[C_AR03-易] 計算陣列中所有元素的立方和

C Version

Howard Guo

















#include <stdio.h>
#include <stdlib.h>

int main(int argc, char *argv[]) {
	int a[6];
	int i=0;
	int total=0;  //要記得設初始值 
	for(i=0;i<6;i++){
		scanf("%d", &a[i]);
	}
	
	for(i=0;i<6;i++){
		total += a[i] * a[i] * a[i];
	}
	printf("%d\n", total);
	return 0;
}

[C_AR04-易] 邊緣偵測

C++ Version

Howard Guo





















































#include <iostream>
#include <cstring>
using namespace std;
int main()
{
    int N, n, m, count_n, count_m; //n上下，m左右
    cin >> N;
    for (int i = 1; i <= N; i++)
    {
        cin >> n >> m;
        int arr[n][m];           //開一個二維陣列
        memset(arr, '0', n * m); //陣列清0
        for (count_n = 0; count_n < n; count_n++)
        {
            for (count_m = 0; count_m < m; count_m++)
            {
                cin >> arr[count_n][count_m];
            }
        }
        //印
        for (count_n = 0; count_n < n; count_n++) //上下
        {
            for (count_m = 0; count_m < m; count_m++) //左右
            {
                //if (count_m == m - 1) //最後一列
                //{
                //    if (arr[count_n][count_m] == 1)
                  //      cout << "0 ";
                    //else
                      //  cout << "_ ";
                //}
                //else
               //{ //一般判斷
                    if (arr[count_n][count_m] == 1)
                    {
                        if (arr[count_n][count_m - 1] == 0 || arr[count_n][count_m + 1] == 0 || arr[count_n - 1][count_m] == 0 || arr[count_n + 1][count_m] == 0)
                        {
                            cout << "0 ";
                        }
                        else
                            cout << "_ ";
                    }
                    else
                        cout << "_ ";
                //}
            }
            cout << "\n";
        }
        if (i != N) //兩圖形間空一行
            cout << "\n";
    }
    return 0;
}

[C_AR05-易] 最少派車數

C Version

風思



























#include <stdio.h>
#include <stdlib.h>

int main(int argc, char *argv[]) {
	int n;
	scanf("%d",&n);
	int start[30]={0};
	int end[30]={0};
	int time[25]={0};
	int i=0,j=0;
	for(i=0;i<n;i++){
		scanf("%d %d",&start[i],&end[i]);
	}
	for(i=0;i<n;i++){
		for(j=start[i];j<end[i];j++){
			time[j]++;
		}
	}
	int max=time[0];
	for(i=1;i<24;i++){
		if(time[i]>max){
			max=time[i];
		}
	}
	printf("%d\n",max);
	return 0;
}

[C_AR09-易] 兩數差值

C++ Version

Howard Guo
arr 的部分也可以用 vector 開動態陣列（但是我還沒學會）






































#include <iostream>
#include <cstring>   //要用memset
#include <algorithm> //要用sort
using namespace std;
int main(int argc, char const *argv[])
{
    string input;
    int arr[20], count, MAX, MIN;
    while (cin >> input)
    {
        MAX = 0, MIN = 0;
        memset(arr, -1, 20);
        count = 0;
        for (int i = 0; i < input.length(); i++)
        { //XXX.length() 取得字串長度
            if (input[i + 1] == ',' || i == input.length() - 1)
            {
                arr[count++] = input[i] - '0';
            }
        }
        //cout << count << endl;    FOR DEBUGGING PURPOSE
        sort(arr, arr + count); //由小排到大

        for (int i = count - 1; i >= 0; i--)
        {
            MAX = MAX * 10 + arr[i];
        }
        //cout << "MAX=" << MAX << endl;    FOR DEBUGGING PURPOSE

        for (int i = 0; i < count; i++)
        {
            MIN = MIN * 10 + arr[i];
        }
        //cout << "MIN=" << MIN << endl;    FOR DEBUGGING PUROPOSE
        cout << MAX - MIN << endl;
    }
    return 0;
}

Java Version

解法一

風思









































import java.util.*;

public class Main {

	public static int[] Bsort(int[] input) { // 泡沫排列法
		for (int i = input.length - 1; i >= 0; i--) {
			for (int j = 0; j < i; j++) {
				if (input[j] > input[j + 1]) {
					int temp = input[j];
					input[j] = input[j + 1];
					input[j + 1] = temp;
				}
			}
		}
		return input;
	}

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		String str = sc.nextLine();
		String tokens[] = (str.split(","));
		int num[] = new int[tokens.length];
		for (int i = 0; i < tokens.length; i++)
			num[i] = Integer.parseInt(tokens[i]);
		Bsort(num); // 泡沫排列
		int max = 0, min = 0;
		// 求min
		for (int i = 0; i < num.length; i++) {
			min += num[i] * (int) Math.pow(10, (double) (num.length - 1 - i));
		}
		// 求max
		for (int i = num.length - 1; i >= 0; i--) {
			max += num[i] * (int) Math.pow(10, (double) (i));
		}
		// System.out.println(max);
		// System.out.println(min);
		System.out.println(max - min);
		sc.close();
	}

}

[C_AR10-中] 新通話費率

Java Version

風思



























































import java.util.*;

//[C_AR10-中] 新通話費率
//2019.07.09;20:41
public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		String[] input = sc.nextLine().split(",");
		double type = Double.parseDouble(input[0]);
		double time = Double.parseDouble(input[1]);
		double fee = 0;
		if (type == 186) {
			if (time * 0.09 < type)
				fee = type;
			else {
				if (time * 0.09 > 2 * type) {
					fee = Math.round(time * 0.09) * 0.8;
				} else {
					fee = Math.round(time * 0.09) * 0.9;
				}
			}
		} else if (type == 386) {
			if (time * 0.08 < type)
				fee = type;
			else {
				if (time * 0.08 > 2 * type) {
					fee = Math.round(time * 0.08) * 0.7;
				} else {
					fee = Math.round(time * 0.08) * 0.8;
				}
			}
		} else if (type == 586) {
			if (time * 0.07 < type)
				fee = type;
			else {
				if (time * 0.07 > 2 * type) {
					fee = Math.round(time * 0.07) * 0.6;
				} else {
					fee = Math.round(time * 0.07) * 0.7;
				}
			}
		} else if (type == 986) {
			if (time * 0.06 < type)
				fee = type;
			else {
				if (time * 0.06 > 2 * type) {
					fee = Math.round(time * 0.06) * 0.5;
				} else {
					fee = Math.round(time * 0.06) * 0.6;
				}
			}
		}
		System.out.printf("%.0f\n", fee);
		sc.close();

	}

}

[C_AR20-易] 檢查數值是否有重複

Java Version

風思
了解Set，不存入重複的資料，再比較輸入的陣列長度。
































import java.util.*;

//[C_AR20-易] 檢查數值是否有重複
//2019.07.20;14:23
public class Main {
	public static boolean checkRespect(String str) {
		Set<String> set = new HashSet<String>();
		String[] tokens = str.split(" ");
		for (String token : tokens)
			set.add(token);
		if (set.size() != tokens.length)
			return true;
		else
			return false;
	}

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt(); // 只是配合題目
		sc.nextLine(); //吸收enter
		String str = sc.nextLine();
		boolean isRespect = checkRespect(str);
		if (isRespect)
			System.out.println(0);
		else
			System.out.println(1);
		sc.close();

	}

}

[C_AR021-易] 成績統計

Java Version

風思
熟習spilt與二維陣列;等待勇者用C寫~

















































import java.util.*;

//[C_AR021-易] 成績統計
// 2019,07,20;15:28
public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		// 輸入學生數
		int n = sc.nextInt();
		sc.nextLine();
		double[][] score = new double[n][3];
		String[][] input = new String[n][3];
		// 一行輸入一個學生的三科分數
		for (int i = 0; i < n; i++) {
			input[i] = sc.nextLine().split(" ");
		}
		// 字串轉數字
		for (int i = 0; i < n; i++) {
			for (int j = 0; j < 3; j++) {
				score[i][j] = Double.parseDouble(input[i][j]);
			}
		}

		double average_all = 0, average_chinese = 0, average_english = 0, average_math = 0;
		// 累加
		for (int i = 0; i < n; i++) {
			for (int j = 0; j < 3; j++) {
				average_all += score[i][j];
				if (j == 0)
					average_chinese += score[i][j];
				if (j == 1)
					average_english += score[i][j];
				if (j == 2)
					average_math += score[i][j];
			}
		}
		// 算出各項平均
		average_all /= (n * 3);
		average_chinese /= n;
		average_english /= n;
		average_math /= n;
		// 輸出
		System.out.printf("%.1f %.1f %.1f %.1f\n", average_all, average_chinese, average_english, average_math);
		sc.close();

	}

}

[C_AR022-易] 字母出現的頻率

C Version

Howard Guo


































#include <stdio.h>
#include <stdlib.h>

/* run this program using the console pauser or add your own getch, system("pause") or input loop */

int main(int argc, char *argv[]) {
	//https://e-tutor.itsa.org.tw/e-Tutor/mod/programming/view.php?id=220
	//[C_AR022-易] 字母出現的頻率
	int result[26] = {0}; //最終輸出，26個字母 初始值=0
	char a[200]; // 輸入值存放處，最大可放200個字元
	int i , temp;  //計數器
	
	scanf("%[^\n]" , &a);  // %[^\n]表示除了\n以外都要讀入，否則scanf遇到空格就會中斷
	
	for(i=0;a[i] !='\0'; i++){  //使a輸出到字串結尾就結束，不用全部印完
		
		if(a[i]>='A' && a[i]<='Z') {  //判斷大寫 
			temp = a[i] - 'A';  //透過ASCII CODE判斷該字元在序列中的順序
			result[temp]++;  //在該位置+1 
		} 
		
		if(a[i]>='a' && a[i]<='z') {  //判斷大寫 
			temp = a[i] - 'a';  //透過ASCII CODE判斷該字元在序列中的順序
			result[temp]++;  //在該位置+1 
		}
	} 
	
	for(i=0;i<26;i++){
		if(i==0) printf("%d", result[i]); //第一個不要加空格 
		else printf(" %d", result[i]);  //之後都要空格 
	}
	printf("\n");
	return 0;
}

[C_AR023-易] 字根與子字串

Java Version

風思


















import java.util.*;

//[C_AR023-易] 字根與子字串
//2019,07,20;16:12
public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		String s = sc.nextLine();
		String t = sc.nextLine();
		if (t.contains(s))
			System.out.println("YES");
		else
			System.out.println("NO");
		sc.close();
	}

}

[C_AR025-易] 計算ASCII字元

C++ Version

Howard Guo
























































#include <iostream>
#include <cstring>
#include <algorithm> //使用sort()
using namespace std;
int main(int argc, char const *argv[])
{
    int num;
    char str[100];
    int arr[100];
    int count = 1, arr_count = 0;
    memset(arr, '0', 100);
    memset(str, ' ', 100);             //陣列清空，這裡不用0是因為測資可能會涵蓋到
    while (cin.getline(str, 99, '\n')) //getline的參數，讀進去str、讀幾個（值必須小於str宣告時的長度）、停止字元(Optional)
    {
        //讀進來以後，要先知道有幾個字元
        for (num = 0; num < 100; num++)
        {
            if (str[num] == ' ')
            {
                num--;
                break;
            }
        }
        sort(str, str + num);    //從小排到大，註：此處用str+num是因為在C/C++中，變數str實際上代表的是該段記憶體位址的首位，這邊是直接操作指標
        reverse(str, str + num); //從大排到小（題目要求）
        arr_count = 0;           //沒有這行的話，第二次輸入後，下面在印出陣列內容時會抓錯值。
        for (int i = 0; i <= num; i++)
        {
            if (str[i + 1] == str[i])
            {
                arr[arr_count] = ++count;
            }
            else
            {
                arr[arr_count] = count;
                count = 1;
                arr_count++;
            }
        }
        arr_count = 0;
        //cout << str << endl;
        //cout << num << endl;
        //cout << arr_count << endl;
        for (int i = 0; i < num; i++)
        {
            if (str[i] != str[i + 1])
            {
                cout << int(str[i]) << " " << arr[arr_count++] << endl;
                //if(arr[arr_count]!=0) cout << endl;
            }
        }
        memset(str, ' ', 100); //陣列清空，準備給下一次迴圈使用
        memset(arr, '0', 100);
    }
    return 0;
}

Java Version

風思





















import java.util.*;  
  
//[C_AR025-易] 計算ASCII字元  
//2019/07/09;22:13  
public class Main {  
  
    public static void main(String[] args) {  
        Scanner sc = new Scanner(System.in);  
        String str = sc.nextLine();  
        int[] result=new int[128];  
        for (int i = 0; i < str.length(); i++) {  
              
            result[str.codePointAt(i)]++; // 返回ASCII數值  
        }  
        for (int i = result.length - 1; i >= 0; i--) {  
            if (result[i] != 0)  
                System.out.printf("%d %d\n", i, result[i]);  
        }  
        sc.close();  
    }  
}

[C_AR029-難] 文字編碼

Java Version

風思
沒比想像的難






































import java.util.*;

//[C_AR029-難] 文字編碼
//2019,08,26;21:40
public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		String inputStr = sc.nextLine();
		int N = inputStr.length();
		int M;
		for (M = 1; M <= 15; M++) {
			if (M * M >= N)
				break;
		}
		String[][] arr = new String[M][M];
		for (int i = 0; i < M; i++)
			Arrays.fill(arr[i], " ");
		String[] tokens = inputStr.split("");
		int index = 0;
		for (int i = 0; i < M; i++) {
			for (int j = 0; j < M; j++) {
				if (index < tokens.length)
					arr[i][j] = tokens[index++];
			}
		}
		StringBuffer strBuf = new StringBuffer();
		for (int i = 0; i < M; i++) {
			for (int j = 0; j < M; j++) {
				strBuf.append(arr[j][i]);
			}
		}
		String result = strBuf.toString();
		System.out.println(result);
		sc.close();
	}

}

[C_AR031-中] 一維矩陣表示二維平面空間

Java Version

風思
有點繁雜




























































import java.util.*;

//[C_AR031-中] 一維矩陣表示二維平面空間
//2019,08,28;22:12
public class Main {
	// N為列數(有幾列)
	public static int indexReturn(int N, int i, int j) {
		// 判斷是否超出邊界,若超出回傳-1
		// 若在邊界內則回傳對應一維矩陣索引值
		if (i < 0 || j >= N || i >= N || j < 0)
			return -1;
		else
			return (i * N + j);
	}

	// 輸入一維矩陣索引值,回傳對應二維矩陣之內容
	public static String getDate(String[][] x, int index) {
		int N = x.length;
		int pi = index / N;
		int pj = index % N;
		if (pi < 0 || pj >= N || pi >= N || pj < 0)
			return "-1";
		else
			return x[pi][pj];

	}

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		int index = sc.nextInt();
		int pi = index / N;
		int pj = index % N;
		String[][] arr = new String[N][N];
		for (int i = 0; i < N; i++)
			for (int j = 0; j < N; j++)
				arr[i][j] = (char) ('A' + i) + "" + (j + 1);
		// 輸出一維矩陣索引
		System.out.print(indexReturn(N, pi - 1, pj - 1) + " ");
		System.out.print(indexReturn(N, pi - 1, pj) + " ");
		System.out.print(indexReturn(N, pi - 1, pj + 1) + " ");
		System.out.print(indexReturn(N, pi, pj + 1) + " ");
		System.out.print(indexReturn(N, pi + 1, pj + 1) + " ");
		System.out.print(indexReturn(N, pi + 1, pj) + " ");
		System.out.print(indexReturn(N, pi + 1, pj - 1) + " ");
		System.out.println(indexReturn(N, pi, pj - 1)+" ");
		// 印出對應資料
		System.out.print(getDate(arr, indexReturn(N, pi - 1, pj - 1)) + " ");
		System.out.print(getDate(arr, indexReturn(N, pi - 1, pj)) + " ");
		System.out.print(getDate(arr, indexReturn(N, pi - 1, pj + 1)) + " ");
		System.out.print(getDate(arr, indexReturn(N, pi, pj + 1)) + " ");
		System.out.print(getDate(arr, indexReturn(N, pi + 1, pj + 1)) + " ");
		System.out.print(getDate(arr, indexReturn(N, pi + 1, pj)) + " ");
		System.out.print(getDate(arr, indexReturn(N, pi + 1, pj - 1)) + " ");
		System.out.println(getDate(arr, indexReturn(N, pi, pj - 1))+" ");
		sc.close();
	}

}

[C_AR34-易] 身分證驗證器

C++ Version

Howard Guo
笨版，用 switch 硬幹，超級笨

























































































#include <iostream>
#include <cstring>
int table(char);
using namespace std;
int main(int argc, char const *argv[])
{
    int X1, X2, N[9], P;

    string id;
    while (getline(cin, id))
    {
        memset(N, 0, 9);
        P = 0;
        X1 = table(id[0]) / 10;
        X2 = table(id[0]) % 10;
        for (int i = 0; i < 9; i++)
        {
            N[i] = id[i + 1] - '0';
        }
        P = X1 + (9 * X2) + (8 * N[0]) + (7 * N[1]) + (6 * N[2]) + (5 * N[3]) + (4 * N[4]) + (3 * N[5]) + (2 * N[6]) + N[7] + N[8];
        if (P % 10 == 0)
        {
            cout << "CORRECT!!!" << endl;
        }
        else
        {
            cout << "WRONG!!!" << endl;
        }
    }
    return 0;
}

int table(char ch)
{
    if (ch == 'A')
        return 10;
    else if (ch == 'B')
        return 11;
    else if (ch == 'C')
        return 12;
    else if (ch == 'D')
        return 13;
    else if (ch == 'E')
        return 14;
    else if (ch == 'F')
        return 15;
    else if (ch == 'G')
        return 16;
    else if (ch == 'H')
        return 17;
    else if (ch == 'J')
        return 18;
    else if (ch == 'K')
        return 19;
    else if (ch == 'L')
        return 20;
    else if (ch == 'M')
        return 21;
    else if (ch == 'N')
        return 22;
    else if (ch == 'P')
        return 23;
    else if (ch == 'Q')
        return 24;
    else if (ch == 'R')
        return 25;
    else if (ch == 'S')
        return 26;
    else if (ch == 'T')
        return 27;
    else if (ch == 'U')
        return 28;
    else if (ch == 'V')
        return 29;
    else if (ch == 'X')
        return 30;
    else if (ch == 'Y')
        return 31;
    else if (ch == 'W')
        return 32;
    else if (ch == 'Z')
        return 33;
    else if (ch == 'I')
        return 34;
    else if (ch == 'O')
        return 35;
    else
        return 0;
}

Java Version

風思
本來也想硬幹，但後來想到有對應的方法~































import java.util.*;

//[C_AR34-易] 身分證驗證器
//2019.07.13;21:46
public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		while (sc.hasNext()) {
			String[] input = sc.nextLine().split("");
			String consult = "ABCDEFGHJKLMNPQRSTUVXYWZIO";
			int[] Code = new int[consult.length()];
			for (int i = 0; i < Code.length; i++)
				Code[i] = i + 10; //賦值
			int code = Code[consult.indexOf(input[0])]; //查表
			int X1 = code / 10;
			int X2 = code % 10;
			int[] N = new int[10]; // N1~N9
			for (int i = 1; i < N.length; i++)
				N[i] = Integer.parseInt(input[i]);
			int P = X1 + 9 * X2 + 8 * N[1] + 7 * N[2] + 6 * N[3] + 5 * N[4] + 4 * N[5] + 3 * N[6] + 2 * N[7] + N[8]
					+ N[9];
			if (P % 10 == 0)
				System.out.println("CORRECT!!!");
			else
				System.out.println("WRONG!!!");
		}
		sc.close();
	}

}

[C_AR35-易] 生肖問題

C Version

風思
























#include <stdio.h>
#define ROW 12
#define COL 10
int count (int x){
    if(x<=2008 && (2008-x)%12==0) return 0; 
    else if(x>2008) return (x-2008)%12;
    else return 12-(2008-x)%12;
}

int main()
{
    char zodic[ROW][COL]={"rat","ox","tiger","rabbit","dragon","snake","horse","sheep","monkey","rooster","dog","pig"};
    int n,i=0,flag=1;
    scanf("%d",&n);
    while(flag==1){
        if(zodic[count(n)][i]=='\0') flag=0; //break;
        else {
            printf("%c",zodic[count(n)][i++]);
        }
    }
    printf("\n");
    return 0;
}

[C_AR36-易] 星座查詢

Java Version

風思














































































import java.util.*;

//[C_AR36-易] 星座查詢
//2019,08,17;17:55
public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		// System.out.println(month+" "+day);
		while (sc.hasNext()) {
			int month = sc.nextInt();
			int day = sc.nextInt();
			if (month == 1) {
				if (day >= 21)
					System.out.println("Aquarius");
				else
					System.out.println("Capricorn");
			} else if (month == 2) {
				if (day >= 19)
					System.out.println("Pisces");
				else
					System.out.println("Aquarius");
			} else if (month == 3) {
				if (day >= 21)
					System.out.println("Aries");
				else
					System.out.println("Pisces");
			} else if (month == 4) {
				if (day >= 21)
					System.out.println("Taurus");
				else
					System.out.println("Aries");
			} else if (month == 5) {
				if (day >= 22)
					System.out.println("Gemini");
				else
					System.out.println("Aries");
			} else if (month == 6) {
				if (day >= 22)
					System.out.println("Cancer");
				else
					System.out.println("Gemini");
			} else if (month == 7) {
				if (day >= 23)
					System.out.println("Leo");
				else
					System.out.println("Cancer");
			} else if (month == 8) {
				if (day >= 24)
					System.out.println("Virgo");
				else
					System.out.println("Leo");
			} else if (month == 9) {
				if (day >= 24)
					System.out.println("Libra");
				else
					System.out.println("Virgo");
			} else if (month == 10) {
				if (day >= 24)
					System.out.println("Scorpio");
				else
					System.out.println("Libra");
			} else if (month == 11) {
				if (day >= 23)
					System.out.println("Sagittarius");
				else
					System.out.println("Scorpio");
			} else if (month == 12) {
				if (day >= 22)
					System.out.println("Capricorn");
				else
					System.out.println("Sagittarius");
			}
		}
		sc.close();
	}

}

[C_AR41-易] 一整數序列所含之整數個數及平均值

Java Version

風思























import java.util.*;

//[C_AR41-易] 一整數序列所含之整數個數及平均值
//2019.07.09;23:14
public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		while (sc.hasNext()) {
			String[] num = sc.nextLine().split(" ");
			int Size = num.length;
			double Average = 0;
			for (int i = 0; i < Size; i++) {
				Average += Double.parseDouble(num[i]);
			}
			Average /= Size;
			System.out.println("Size: " + Size);
			System.out.printf("Average: %.3f\n", Average);
		}
		sc.close();
	}

}

[C_AR42-易] 過半元素

Java Version

風思












































import java.util.*;

//[C_AR42-易] 過半元素
//2019.07.13;19:58
public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		while (sc.hasNext()) {
			String input = sc.nextLine();
			String[] nums = input.split(" ");
			String[] control = input.split(" "); // 對照組
			Double N = nums.length / 2.0;
			int[] result = new int[nums.length];
			for (int i = 0; i < nums.length; i++) {
				for (int j = i; j < nums.length && (!nums[j].equals("iscount")); j++) {
					if (nums[i].equals(nums[j]) && (!nums[j].equals("iscount"))) {
						result[i]++;
						if (i != j)
							nums[j] = "iscount";
					}

				}
				nums[i] = "iscount";
			}
			boolean isfind = false;
			int index = 0;
			for (int i = 0; i < result.length; i++) {
				if (result[i] > N) {
					isfind = true;
					index = i;
					break;
				}
			}
			if (isfind)
				System.out.println(control[index]);
			else
				System.out.println("NO");

		}
		sc.close();
	}

}

[C_AR43-易] 以遞迴函數計算冪次

C Version

風思
















#include <stdio.h>
unsigned long long intPower(int m,int n){
    if(n==0) return 1;
    else if(n==1) return m;
    else return (m*intPower(m,n-1));
}
int main()
{
    int m,n;
    while(scanf("%d %d",&m,&n)!=EOF){
        printf("%ld",intPower(m,n));
        printf("\n");
    }
    return 0;
}

[C_AR44-易] 迴文問題

C Version

風思











































#include <stdio.h>
#include <stdlib.h>

int main(int argc, char *argv[]) {
	char a[80];
	
	scanf("%s",&a);
	
	int size = 0 ;
	int flag = 1;
	int i ;
	
	for(i=0;i<80;i++){
		if(a[i]=='\0'){
			break;
			
		
		}
		size++;
		
	}
	

	//printf("%d\n",size);
	
	
	for(i=0;i<size/2;i++){
		if(a[i] != a[size-1-i]){
			printf("NO\n");
			flag =0;
			break;
			
		}
		
		
		
	}
	if(flag==1) printf("YES\n");
	
	
	
	return 0;
}

[C_AR45-易] 輸入字串算字元

Java Version

風思
就是這麼簡單













import java.util.*;

//[C_AR45-易] 輸入字串算字元
//2019,7,09;22:20
public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		System.out.printf("There are %d characters\n", sc.nextLine().length());
		sc.close();
	}

}

[C_AR47-易] 利用指標傳遞陣列到函數

Java Version

風思
有空再用C寫吧~






















import java.util.*;

//[C_AR47-易] 利用指標傳遞陣列到函數
//2019.07.09;22:57
public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		while (sc.hasNext()) {
			String[] num = sc.nextLine().split(" ");
			for (int i = num.length - 1; i >= 0; i--) {
				if (i != 0)
					System.out.print(num[i] + " ");
				else
					System.out.println(num[i]);
			}
		}
		sc.close();
	}

}

[C_AR48-易] 數字加密

C Version

風思























#include <stdio.h>
int change (int arr[],int size){
    int i;
    for(i=0;i<size;i++) arr[i]=(arr[i]+7)%10;
    int temp =arr[0];
    arr[0]=arr[2];
    arr[2]=temp;
    int temp2=arr[1];
    arr[1]=arr[3];
    arr[3]=temp2;
}

int main()
{
    int key [4]={0};
    int i=0;
    for (i=0;i<4;i++)scanf("%1d",&key[i]);
    change(key,4);
    for(i=0;i<4;i++) printf("%d",key[i]);
    printf("\n");
    return 0;
}

[C_AR57-易] 刪除重複資料

此題用 C++ 的 cin 跟 getline 需要注意這個觀念

Syntax	Example	Reference
# Header	Header	基本排版
- Unordered List	Unordered List
1. Ordered List	Ordered List
- [ ] Todo List	Todo List
> Blockquote	Blockquote
Bold font	Bold font
Italics font	Italics font
~~Strikethrough~~	~~Strikethrough~~
19^th^	19^th
H~2~O	H₂O
++Inserted text++	Inserted text
==Marked text==	Marked text
[link text](https:// "title")	Link
![image alt](https:// "title")	Image
`Code`	`Code`	在筆記中貼入程式碼
```javascript var i = 0; ```	`var i = 0;`	在筆記中貼入程式碼
:smile:		Emoji list
{%youtube youtube_id %}	Externals
$L^aT_eX$	L^aT_eX
:::info This is a alert area. :::	This is a alert area.