# ITSA E-tutor Code Exchange
# READ ME!
**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）**
1. 首先找到程式練習的題庫，沒有的話用 h2 （兩個#）開新的區塊。
2. 使用 h3 （三個#） 貼上題名(再放置連結)，E-tutor格式如下。
3. 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。
4. 可標示程式碼提供者。
5. 本頁面使用 C 寫的 Code 在格式部分請選擇「cpp=」。
6. 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。
7. 程式碼力求可讀性高(因為看到自己的黑歷史)。
8. 善用註解，另可額外補充解決此題應知資訊。
9. 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。
10. 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。
```java=
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
> [name=Howard Guo][color=#4287f5]
---
> 每天都要記得 Coding，C、C++、Java至少要碰其中一個!
> [time=Sat, Jul 20, 2019 1:16 AM]
> [name=風思] [color=#7cfc00]
---
## [Mathematics I](https://e-tutor.itsa.org.tw/e-Tutor/course/view.php?id=284)
### [[C_MM01-易] 計算梯型面積](https://e-tutor.itsa.org.tw/e-Tutor/mod/programming/view.php?id=6855)
#### C Version
> [name=Howard Guo][color=#4287f5]
```cpp=
#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;
}
```
> [name=風思]
```cpp=
#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
> [name=AfreecaKing]
```java=
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](https://e-tutor.itsa.org.tw/e-Tutor/mod/programming/view.php?id=6858)
#### Java Version
> [name=AfreecaKing]
```java=
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](https://e-tutor.itsa.org.tw/e-Tutor/mod/programming/view.php?id=6861)
#### Java Version
> [name=AfreecaKing]
```java=
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](https://e-tutor.itsa.org.tw/e-Tutor/mod/programming/view.php?id=6864)
#### Java Version
> [name=AfreecaKing]
> 注意餘數不能是負的
```java=
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](https://e-tutor.itsa.org.tw/e-Tutor/mod/programming/view.php?id=6867)
#### C Version
> [name=風思]
>當初卡很久在四捨五入的問題
```cpp=
#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
> [name=風思]
> 相同的格式化輸出，java四捨五入；C五捨六入
```java=
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
> [name=風思]
> 當初寫C要是知道這招，就好啦!
```java=
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](https://e-tutor.itsa.org.tw/e-Tutor/mod/programming/view.php?id=6870)
#### Java Version
> [name=AfreecaKing]
```java=
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](https://e-tutor.itsa.org.tw/e-Tutor/mod/programming/view.php?id=6873)
#### Java Version
> [name=AfreecaKing]
```java=
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](https://e-tutor.itsa.org.tw/e-Tutor/mod/programming/view.php?id=6876)
#### Java Version
> [name=Afreecaking]
```java=
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](https://e-tutor.itsa.org.tw/e-Tutor/mod/programming/view.php?id=6879)
#### Java Version
> [name=AfreecaKing]
```java=
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](https://e-tutor.itsa.org.tw/e-Tutor/mod/programming/view.php?id=6882)
#### Java Version
> [name=AfreecaKing]
> 強烈建議不要用C寫，浮點數會有誤差
```java=
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
> [name=AfreecaKing]
> 用round來處理浮點數誤差，很久以前寫的可讀性不高
```cpp=
#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](https://e-tutor.itsa.org.tw/e-Tutor/mod/programming/view.php?a=2923)
#### C Version
> [name=風思]
> 天罰版，可讀性幾乎為0
```cpp=
#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
> [name=風思]
> 好多了
```cpp=
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
> [name=AfreecaKing]
```java=
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](https://e-tutor.itsa.org.tw/e-Tutor/mod/programming/view.php?id=6888)
#### Java Version
> [name=AfreecaKing]
```java=
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](https://e-tutor.itsa.org.tw/e-Tutor/mod/programming/view.php?id=6891)
#### Java Version
> [name=AfreecaKing]
> 將分鐘換成秒來計算
```java=
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](https://e-tutor.itsa.org.tw/e-Tutor/mod/programming/view.php?id=6894)
#### Java Version
> [name=AfreecaKing]
```java=
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](https://e-tutor.itsa.org.tw/e-Tutor/mod/programming/view.php?id=6897)
#### Java Version
> [name=AfreecaKing]
```java=
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](https://e-tutor.itsa.org.tw/e-Tutor/mod/programming/view.php?id=6900)
#### Java Version
> [name=AfreecaKing]
```java=
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](https://e-tutor.itsa.org.tw/e-Tutor/mod/programming/view.php?id=6903)
#### Java Version
> [name=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
```java=
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](https://e-tutor.itsa.org.tw/e-Tutor/mod/programming/view.php?a=2930)
#### Java Version
> [name=風思]
> ~n為n的[ones' complement](https://zh.wikipedia.org/wiki/%E4%B8%80%E8%A3%9C%E6%95%B8)（一的補數）
```java=
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](https://e-tutor.itsa.org.tw/e-Tutor/mod/programming/view.php?id=6909)
#### Java Version
> [name=AfreecaKing]
```java=
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](https://e-tutor.itsa.org.tw/e-Tutor/mod/programming/view.php?id=6949)
#### C++ Version
> 這題也是用笨笨 switch 硬幹， There should be a better solution even it's a simple problem.
> [name=Howard Guo][color=#4287f5]
```cpp=
#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
> [name=風思]
嘿嘿嘿，比↑少了粉多break跟輸出~
```cpp=
#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
> [name=風思]
```java=
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](https://e-tutor.itsa.org.tw/e-Tutor/mod/programming/view.php?id=6951)
#### C++ Version
> [name=Howard Guo][color=#4287f5]
```cpp=
#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
> [name=風思]
```cpp=
#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](https://e-tutor.itsa.org.tw/e-Tutor/mod/programming/view.php?id=6953)
#### C++ Version
> [name=Howard Guo][color=#4287f5]
```cpp=
#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
> [name=風思]
```cpp=
#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
> [name=風思]
```cpp=
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](https://e-tutor.itsa.org.tw/e-Tutor/mod/programming/view.php?id=6955)
#### C++ Version
> [name=Howard Guo][color=#4287f5]
```cpp=
#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
> [name=風思]
```cpp=
#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](https://e-tutor.itsa.org.tw/e-Tutor/mod/programming/view.php?id=6957)
#### C++ Version
> [name=Howard Guo][color=#4287f5]
```cpp=
#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
> [name=風思]
```cpp=
#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](https://e-tutor.itsa.org.tw/e-Tutor/mod/programming/view.php?id=6965)
#### C Version
> [name=風思]
> scanf("%[^\n]",&N)為可輸入空白之字串表示方法
```cpp=
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](https://e-tutor.itsa.org.tw/e-Tutor/mod/programming/view.php?id=6976)
#### C++ Version
> [name=Howard Guo][color=#4287f5]
```cpp=
#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
> [name=風思]
```cpp=
#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](https://e-tutor.itsa.org.tw/e-Tutor/course/view.php?id=63)
### [[C_AR01-易] 一維陣列反轉 I](https://e-tutor.itsa.org.tw/e-Tutor/mod/programming/view.php?id=674)
#### C++ Version
> [name=Howard Guo][color=#4287f5]
```cpp=
#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
> [name=風思]
> 未來你會看到很多spilt();
```java=
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](https://e-tutor.itsa.org.tw/e-Tutor/mod/programming/view.php?id=1289)
#### C Version
> [name=風思]
```cpp=
#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;
}
```
> [name=Howard Guo][color=#4287f5]
```cpp=
#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](https://e-tutor.itsa.org.tw/e-Tutor/mod/programming/view.php?id=1291)
#### C Version
> [name=Howard Guo][color=#4287f5]
```cpp=
#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](https://e-tutor.itsa.org.tw/e-Tutor/mod/programming/view.php?id=1703)
#### C++ Version
> [name=Howard Guo][color=#4287f5]
```cpp=
#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](https://e-tutor.itsa.org.tw/e-Tutor/mod/programming/view.php?id=1708)
#### C Version
> [name=風思]
```cpp=
#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](https://e-tutor.itsa.org.tw/e-Tutor/mod/programming/view.php?a=1006)
#### C++ Version
> [name=Howard Guo][color=#4287f5]
> arr 的部分也可以用 vector 開動態陣列（但是我還沒學會）
```cpp=
#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
> [name=風思]
```java=
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](https://e-tutor.itsa.org.tw/e-Tutor/mod/programming/view.php?id=1912)
#### Java Version
> [name=風思]
```java=
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](https://e-tutor.itsa.org.tw/e-Tutor/mod/programming/view.php?a=1113)
#### Java Version
> [name=風思]
> 了解Set，不存入重複的資料，再比較輸入的陣列長度。
```java=
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](https://e-tutor.itsa.org.tw/e-Tutor/mod/programming/view.php?a=1122)
#### Java Version
> [name=風思]
> 熟習spilt與二維陣列;等待勇者用C寫~
```java=
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](https://e-tutor.itsa.org.tw/e-Tutor/mod/programming/view.php?id=2200)
#### C Version
> [name=Howard Guo] [color=#4287f5]
```cpp=
#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](https://e-tutor.itsa.org.tw/e-Tutor/mod/programming/view.php?a=1125)
#### Java Version
> [name=風思]
```java=
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](https://e-tutor.itsa.org.tw/e-Tutor/mod/programming/view.php?id=2435)
#### C++ Version
> [name=Howard Guo][color=#4287f5]
```cpp=
#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
> [name=風思]
```java=
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](https://e-tutor.itsa.org.tw/e-Tutor/mod/programming/view.php?id=2454)
#### Java Version
> [name=風思]
> 沒比想像的難
```java=
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](https://e-tutor.itsa.org.tw/e-Tutor/mod/programming/view.php?id=2606)
#### Java Version
> [name=風思]
> 有點繁雜
```java=
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](https://e-tutor.itsa.org.tw/e-Tutor/mod/programming/view.php?id=2638)
#### C++ Version
> [name=Howard Guo]
> 笨版，用 switch 硬幹，超級笨
> [color=#4287f5]
```cpp=
#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
> [name=風思]
> 本來也想硬幹，但後來想到有對應的方法~
``` java=
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](https://e-tutor.itsa.org.tw/e-Tutor/mod/programming/view.php?id=2641)
#### C Version
> [name=風思]
```cpp=
#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](https://e-tutor.itsa.org.tw/e-Tutor/mod/programming/view.php?id=2644)
#### Java Version
> [name=風思]
```java=
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](https://e-tutor.itsa.org.tw/e-Tutor/mod/programming/view.php?id=3039)
#### Java Version
> [name=風思]
```java=
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](https://e-tutor.itsa.org.tw/e-Tutor/mod/programming/view.php?id=3042)
#### Java Version
> [name=風思]
```java=
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](https://e-tutor.itsa.org.tw/e-Tutor/mod/programming/view.php?id=3045)
#### C Version
> [name=風思]
```cpp=
#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](https://e-tutor.itsa.org.tw/e-Tutor/mod/programming/view.php?id=3067)
#### C Version
> [name=風思]
```cpp=
#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](https://e-tutor.itsa.org.tw/e-Tutor/mod/programming/view.php?a=1444)
#### Java Version
> [name=風思]
> 就是這麼簡單
```java=
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](https://e-tutor.itsa.org.tw/e-Tutor/mod/programming/view.php?a=1446)
#### Java Version
> [name=風思]
> 有空再用C寫吧~
```java=
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](https://e-tutor.itsa.org.tw/e-Tutor/mod/programming/view.php?a=1447)
#### C Version
> [name=風思]
```cpp=
#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](https://e-tutor.itsa.org.tw/e-Tutor/mod/programming/view.php?id=3516)
此題用 C++ 的 cin 跟 getline 需要注意 [這個觀念](http://justimchung.blogspot.com/2016/11/c-cin-getline.html)
Google Drive
Gist
Clipboard
Sign in with Wallet
