# ITSA E-tutor 程式碼參考(數學1)
{%hackmd BJrTq20hE %}
## [數學 I(Mathematics)](https://e-tutor.itsa.org.tw/e-Tutor/mod/programming/index.php?id=284)
> [name=風思] 新增 [Python版本](https://hackmd.io/GsvoEIfMQAy6qtg-4gCDSA) <- 比較漂亮
> ~~看到以前寫的，感覺自己好菜喔，連i都因為學C的緣故沒直接在for迴圈裡宣告。~~
> [time=Wed, Jun 30, 2021 10:20 PM]
---
> [name=風思]有幾題難題以井字及紅字標誌，可先跳過，其中涉及較複雜的演算法。
> [time=Fri, Aug 21, 2020 9:32 PM]
---
> [name=風思] 就當參考吧~反正我C++也剛學1個月（x
> 程式碼力求簡潔，讓別人看懂也是不容易的。
> 結束43題~
> [time=Sat, Dec 14, 2019 10:47 PM]
---
> [TOC]
### [C_MM01-易] 計算梯型面積
#### C++:
```cpp=
#include <iostream>
#include <iomanip>
using namespace std;
int main()
{
//宣告三數分別代表上底、下底、高
int a,b,c;
//輸入
cin >> a >> b >> c;
//計算梯形面積
double area = ((a+b)*c)/2.0;
//四捨五入取到小數點第一位
cout << fixed << setprecision(1) << "Trapezoid area:" << area << endl;
return 0;
}
```
### [C_MM02-易] 計算三角形面積
#### C++:
```cpp=
#include <iostream>
#include <iomanip>
using namespace std;
int main()
{
//宣告兩數分別代表底、高
int a,b;
//輸入
cin >> a >> b;
//計算面積
double area = (a*b)/2.0;
//四捨五入取到小數點第一位
cout << fixed << setprecision(1) << area << endl;
return 0;
}
```
### [C_MM03-易] 兩數總和
#### C++:
```cpp=
#include <iostream>
using namespace std;
int main()
{
//宣告兩數
int a,b;
//重複輸入
while(cin >> a >> b){
cout << a+b << endl;
}
return 0;
}
```
### [C_MM04-易] 計算總和、乘積、差、商和餘數
#### C++:
```cpp=
#include <iostream>
using namespace std;
int main()
{
int a,b;
cin >> a >> b;
cout << a << "+" << b << "=" << a+b << endl;
cout << a << "*" << b << "=" << a*b << endl;
cout << a << "-" << b << "=" << a-b << endl;
//餘數>=0
int c=a%b;
if(c<0)
cout << a << "/" << b << "=" << (a/b)-1 << "..." << c+b <<endl;
else
cout << a << "/" << b << "=" << (a/b) << "..." << c <<endl;
return 0;
}
```
### [C_MM05-易] 計算正方形面積
#### C++:
```cpp=
#include <iostream>
#include <iomanip>
#include <cmath>
using namespace std;
int main()
{
double a;
cin >> a ;
//round函式四捨五入到個數位，故乘10後round，再除10.0回來
cout << fixed << setprecision(1) << round(a*a*10)/10.0 << endl;
return 0;
}
```
### [C_MM06-易] 英哩轉公里
#### C++:
```cpp=
#include <iostream>
#include <iomanip>
#include <cmath>
using namespace std;
int main()
{
int mile;
cin >> mile;
double km = mile * 1.6;
cout << fixed << setprecision(1) << km << endl;
return 0;
}
```
### [C_MM07-易] 計算平方值與立方值
#### C++:
```cpp=
#include <iostream>
using namespace std;
int main()
{
int x;
cin >> x;
cout << x << " " << x*x << " " << x*x*x << endl;
return 0;
}
```
### [C_MM08-易] 計算兩數和的平方值
#### C++:
```cpp=
#include <iostream>
using namespace std;
int main()
{
int x,y,z;
cin >> x >> y;
z=x+y;
cout << z*z << endl;
return 0;
}
```
### [C_MM09-易] 計算 i 次方的值
#### C++:
##### <特解>解法一:
```cpp=
#include <iostream>
using namespace std;
int main()
{
int i,result=1;
cin >> i;
if(i>31)
cout << "Value of more than 31" << endl;
else
{
//位移運算元，每往左即乘2
result = result << i;
cout << result << endl;
}
return 0;
}
```
##### <直觀>解法二:
```cpp=
#include <iostream>
using namespace std;
int main()
{
int i,j,result=1;
cin >> i;
if(i>31)
cout << "Value of more than 31" << endl;
else
{
//使用迴圈
for(j=1; j<=i; j++)
result *=2;
cout << result << endl;
}
return 0;
}
```
#### <直觀>解法三:
```cpp=
#include <iostream>
#include <cmath>
using namespace std;
int main()
{
int i,result;
cin >> i;
if(i>31)
cout << "Value of more than 31" << endl;
else
{
//使用cmath中的pow函數
result=pow(2,i);
cout << result << endl;
}
return 0;
}
```
### [C_MM10-易] 攝氏溫度轉華式溫度
#### C++:
```cpp=
#include <iostream>
#include <iomanip>
using namespace std;
int main()
{
double c,f;
cin >> c;
f=c*(9.0/5.0)+32;
cout << fixed << setprecision(1) << f << endl;
return 0;
}
```
### [C_MM11-易] 購票計算
#### C++:
```cpp=
#include <iostream>
using namespace std;
int main()
{
int n,n10,n5,n1;
cin >> n;
//%表取餘數，而int間相除仍為整數
n10=n/10;
n5=(n%10)/5;
n1=n%5;
cout << "NT10=" << n10 << endl;
cout << "NT5=" << n5 << endl;
cout << "NT1=" << n1 << endl;
return 0;
}
```
### [C_MM12-易] 相遇時間計算
#### C++:
```cpp=
#include <iostream>
#include <cmath>
using namespace std;
int main()
{
//dist表distance,t表秒數
double dist,t;
cin >> dist;
//換成公分
dist *=100;
t=dist/(100-30*2.54);
//無條件進位
cout << ceil(t) << endl;
return 0;
}
```
### [C_MM13-易] 停車費計算
#### C++:
```cpp=
#include <iostream>
using namespace std;
int main()
{
int h1,h2,m1,m2,t,sum=0;;
cin >> h1 >> m1 >> h2 >> m2;
t = 60*(h2-h1)+(m2-m1);
if(t<=120 and t>=30)
sum=(t/30)*30;
else if(t>120 and t<=240)
sum=((t-120)/30)*40+120;
else if(t>240)
sum=((t-240)/30)*60+280;
cout << sum << endl;
return 0;
}
```
### [C_MM14-易] 計算時間的組合
#### C++:
```cpp=
#include <iostream>
using namespace std;
int main()
{
int s,days,hours,minutes,seconds;
cin >> s;
days = s/86400;
hours = (s%86400)/3600;
minutes = (s%3600)/60;
seconds = s%60;
cout << days << " days" << endl;
cout << hours << " hours" << endl;
cout << minutes << " minutes" << endl;
cout << seconds << " seconds" << endl;
return 0;
}
```
### [C_MM15-易] 判斷座標是否在正方形的範圍內
#### C++:
```cpp=
#include <iostream>
using namespace std;
int main()
{
int x,y;
cin >> x >> y;
if(x>=0 and x<=100 and y>=0 and y<=100)
cout << "inside" << endl;
else
cout << "outside" << endl;
return 0;
}
```
### [C_MM16-易] 判斷座標是否在圓形的範圍內
#### C++:
```cpp=
#include <iostream>
using namespace std;
int main()
{
int x,y;
cin >> x >> y;
if(x*x<=10000 and y*y<=10000)
cout << "inside" << endl;
else
cout << "outside" << endl;
return 0;
}
```
### [C_MM17-易] 求最大公因數
#### C++:
#### <直觀>解法一(迴圈):
取小數開始，較有效率。
```cpp=
#include <iostream>
using namespace std;
int main()
{
int x,y,temp,i;
cin >> x >> y;
//設定x為較大的數
if(y>x)
{
temp=x;
x=y;
y=temp;
}
for(i=y; i>=1; i--)
{
if(x%i==0 and y%i==0)
{
cout << i << endl;
break;
}
}
return 0;
}
```
#### <推薦>解法二(遞迴):
輾轉相除法(gcd)滿足交換律，gcd(x,y)=gcd(y,x)，不必管x與y的大小關係。
```cpp=
#include <iostream>
using namespace std;
int gcd(int x,int y)
{
if(x%y==0)
return y;
else
return gcd(y,x%y);
}
int main()
{
int x,y,result;
cin >> x >> y;
result=gcd(x,y);
cout << result << endl;
return 0;
}
```
### [C_MM18-易] 十進制轉二進制
#### C++:
"~"表[取一補數(ones' complement)](https://zh.wikipedia.org/wiki/%E4%B8%80%E8%A3%9C%E6%95%B8)，人腦處理負數轉[二進位](https://zh.wikipedia.org/wiki/%E4%BA%8C%E8%BF%9B%E5%88%B6)時用[二補數(2's complement)](https://zh.wikipedia.org/wiki/%E4%BA%8C%E8%A3%9C%E6%95%B8)，但我們用陣列操作起來不方便，所以我便有了以下想法:
> [name=風思] ~~以下開放膜拜:~~
> [color=#5165c6]
```cpp=
#include <iostream>
using namespace std;
int main()
{
int x,i;
int result[8];
cin >> x;
bool isneg = (x<0);
//負數的話則取一的補數，轉成正數處理
if(isneg)
x=~x;
//實踐二進位，倒置放入陣列
for(i=0; i<8; i++)
{
result[7-i]=x%2;
x/=2;
}
//再轉回來
if(isneg)
{
for(i=0; i<8; i++)
{
if(result[i]==0)
result[i]=1;
else
result[i]=0;
}
}
//輸出，最後換行
for(i=0; i<8; i++)
cout << result[i];
cout << "\n";
return 0;
}
```
### [C_MM19-易] 電話費計算
#### C++:
```cpp=
#include <iostream>
#include <iomanip>
using namespace std;
int main()
{
int t;
double sum=0;
cin >> t;
if(t<=800)
sum=t*0.9;
else if(t>800 and t<1500)
sum=(t*0.9)*0.9;
else if(t>=1500)
sum=(t*0.9)*0.79;
cout << fixed << setprecision(1) << sum << endl;
return 0;
}
```
### [C_MM20-易] 十進位轉十六進位
#### C++:
hex表16進位輸出，uppercase表大寫輸出。
```cpp=
#include <iostream>
#include <iomanip>
using namespace std;
int main()
{
int x;
cin >> x;
cout << hex << uppercase << x << endl;
return 0;
}
```
### [C_MM21-易] 算階乘
#### C++:
```cpp=
#include <iostream>
using namespace std;
int main()
{
int n,i;
//int型態會溢位
long sum=1;
cin >> n;
for(i=1; i<=n; i++)
sum*=i;
cout << sum << endl;
return 0;
}
```
### [C_MM24-易] 計算薪水
#### C++:
```cpp=
#include <iostream>
#include <iomanip>
using namespace std;
int main()
{
int hour,pay;
double salary=0;
cin >> hour >> pay;
if(hour<=60)
salary=hour*pay;
else if(hour>60 and hour<121)
salary=(60*pay)+(hour-60)*pay*1.33;
else if(hour>=121)
salary=(60*pay*2.33)+(hour-120)*pay*1.66;
cout << fixed << setprecision(1) << salary << endl;
return 0;
}
```
### [C_MM25-易] 計算正整數被3整除之數值之總和
#### C++:
```cpp=
#include <iostream>
using namespace std;
int main()
{
int N,i,sum=0;
cin >> N;
for(i=1; i<=N; i++)
{
if(i%3==0)
sum+=i;
}
cout << sum << endl;
return 0;
}
```
### [C_MM26-易] 輸出 1*1、2*2、...、N*N之結果
#### C++:
```cpp=
#include <iostream>
using namespace std;
int main()
{
int N,i;
cin >> N;
for(i=1; i<=N; i++)
cout << i << "*" << i << "=" << i*i << endl;
return 0;
}
```
### [C_MM27-易] 計算兩整數間所有整數的總和
#### C++:
#### <特解>解法一(純數):
abs取絕對值
```cpp=
#include <iostream>
#include <cmath>
using namespace std;
int main()
{
int a,b;
cin >> a >> b;
int sum= ((a+b)*(abs(a-b)+1))/2;
cout << sum << endl;
return 0;
}
```
#### <直觀>解法二(迴圈):
```cpp=
#include <iostream>
using namespace std;
int main()
{
int a,b,i,temp,sum=0;
cin >> a >> b;
//設定b為大數
if(a>b)
{
temp=a;
a=b;
b=temp;
}
for(i=a; i<=b; i++)
sum+=i;
cout << sum << endl;
return 0;
}
```
### [C_MM28-易] 計算1到N之間屬於5和7的倍數
#### C++:
最後一個不能有空格，且記得換行。
```cpp=
#include <iostream>
using namespace std;
int main()
{
int N,i;
cin >> N;
for(i=1; i<=N; i++)
{
if(i%35==0)
{
if(i==35)
cout << i;
else
cout << " " << i;
}
}
cout << "\n";
return 0;
}
```
### [C_MM29-易] 最大質數問題
#### C++:
```cpp=
#include <iostream>
using namespace std;
int main()
{
int x,i,j;
bool isprime;
cin >> x;
for(i=x-1; i>=2; i--)
{
isprime=true;
for(j=2; j<i; j++)
{
if(i%j==0)
{
isprime=false;
break;
}
}
if(isprime)
{
cout << i << endl;
break;
}
}
return 0;
}
```
### [C_MM30-易] 質數判別
#### C++:
```cpp=
#include <iostream>
using namespace std;
int main()
{
int x,i;
bool isprime=true;
cin >> x;
for(i=2; i<x; i++)
{
if(x%i==0)
{
isprime=false;
break;
}
}
if(isprime)
cout << "YES" << endl;
else
cout << "NO" << endl;
return 0;
}
```
### [C_MM31-易] 計算1~N內能被2跟3整除，但不能被12整除的整數總和
#### C++:
```cpp=
#include <iostream>
using namespace std;
int main()
{
int N,i,sum=0;
cin >> N;
for(i=1; i<=N; i++)
{
if(i%6==0 and i%12!=0)
sum+=i;
}
cout << sum << endl;
return 0;
}
```
### [C_MM32-易] Armstrong數
#### C++:
```cpp=
#include <iostream>
#include <cmath>
using namespace std;
int main()
{
int x,a,b,c;
cin >> x;
a=x/100;
b=(x%100)/10;
c=x%10;
bool isArmstrong=(pow(a,3)+pow(b,3)+pow(c,3))==x;
if(isArmstrong)
cout << "Yes" << endl;
else
cout << "No" << endl;
return 0;
}
```
### [C_MM33-易] 找1~N的完美數
#### C++:
第一個完美數為6，且完美數必為偶數。
```cpp=
#include <iostream>
using namespace std;
int main()
{
int N,i,j,sum;
cin >> N;
for(i=6; i<=N; i+=2)
{
sum=0;
for(j=1; j<i; j++)
{
if(i%j==0)
sum+=j;
}
if(sum==i)
{
if(i==6)
cout << i;
else
cout << " " << i;
}
}
cout << "\n";
return 0;
}
```
### [C_MM34-易] 因數問題
#### C++:
```cpp=
#include <iostream>
using namespace std;
int main()
{
int x,i;
cin >>x;
for(i=1; i<=x; i++)
{
if(x%i==0)
{
if(i==1)
cout << i;
else
cout << " " << i;
}
}
cout << "\n";
return 0;
}
```
### [C_MM35-易] 平、閏年判定
#### C++:
```cpp=
#include <iostream>
using namespace std;
int main()
{
int year;
cin >> year;
if(year%4==0)
{
if(year%400==0)
cout << "Bissextile Year" << endl;
else
cout << "Common Year" << endl;
}
else
cout << "Common Year" << endl;
return 0;
}
```
### [C_MM36-易] 季節判定
#### C++:
```cpp=
#include <iostream>
using namespace std;
int main()
{
int month;
cin >> month;
switch (month)
{
case 3:
case 4:
case 5:
cout << "Spring" << endl;
break;
case 6:
case 7:
case 8:
cout << "Summer" << endl;
break;
case 9:
case 10:
case 11:
cout << "Autumn" << endl;
break;
case 12:
case 1:
case 2:
cout << "Winter" << endl;
break;
}
return 0;
}
```
### [C_MM37-易] 判斷座標位於何處
#### C++:
```cpp=
#include <iostream>
using namespace std;
int main()
{
int x,y;
cin >> x >> y;
if(x==0 and y==0)
cout << "Origin" << endl;
else if(x!=0 and y==0)
cout << "x-axis" << endl;
else if(x==0 and y!=0)
cout << "y-axis" << endl;
else if(x>0 and y>0)
cout << "1st Quadrant" << endl;
else if(x<0 and y>0)
cout << "2nd Quadrant" << endl;
else if(x<0 and y<0)
cout << "3rd Quadrant" << endl;
else if(x>0 and y<0)
cout << "4th Quadrant" << endl;
return 0;
}
```
### [C_MM38-易] 判斷3整數是否能構成三角形之三邊長
#### C++:
```cpp=
#include <iostream>
using namespace std;
int main()
{
int x,y,z;
cin >> x >> y >> z;
if(x+y>z and x+z>y and y+z>x)
cout << "fit" << endl;
else
cout << "unfit" << endl;
return 0;
}
```
### [C_MM39-易] 判斷是何種三角形
#### C++:
```cpp=
#include <iostream>
using namespace std;
int main()
{
int a,b,c,a2,b2,c2;
cin >> a >> b >> c;
a2=a*a;
b2=b*b;
c2=c*c;
if(a+b>c and a+c>b and b+c>a)
{
if(a2+b2==c2 or a2+c2==b2 or b2+c2==a2)
cout << "Right Triangle" << endl;
else if(a2+b2<c2 or a2+c2<b2 or b2+c2<a2)
cout << "Obtuse Triangle" << endl;
else if(a2+b2>c2 and a2+c2>b2 and b2+c2>a2)
cout << "Acute Triangle" << endl;
}
else
cout << "Not Triangle" << endl;
return 0;
}
```
### [C_MM40-易] 1~N之間的總和
#### C++:
```cpp=
#include <iostream>
using namespace std;
int main()
{
int N,i,sum=0;
cin >> N;
for(i=1; i<=N; i++)
{
if(i==1)
cout << i << " ";
else
cout << "+ " << i << " ";
sum+=i;
}
cout << "= " << sum << endl;
return 0;
}
```
### [C_MM42-中] 求(-1)^(n+1)*[1/(2n-1)]的和
#### C++:
```cpp=
#include <iostream>
#include <cmath>
#include <iomanip>
using namespace std;
int main()
{
int n,i;
double S=0;
cin >> n;
for(i=1; i<=n; i++)
S+=pow(-1.0,(i+1)*1.0)*1/(2*i-1);
cout << fixed << setprecision(3) << S << endl;
return 0;
}
```
### [C_MM44-易] The Numbers
#### C++:
```cpp=
#include <iostream>
using namespace std;
int main()
{
char N[2];
char M[8];
int count=0;
cin >> N;
cin >> M;
for(int i=0; i<=6; i++)
{
if(M[i]==N[0] and M[i+1]==N[1])
count++;
}
cout << count << endl;
return 0;
}
```
### <font color="#f00">#[C_MM45-中] 分禮物</font>
#### Java:
```java=
import java.util.*;
class P {
int n;
int[] a;
int[] b;
int k;
Human[] arr;
StringBuffer strBuf;
P(int n) {
this.n = n;
this.k = 0;
strBuf = new StringBuffer();
a = new int[n + 1];
b = new int[n + 1];
arr = new Human[n + 1];
}
void dfs(int step) {
// 完成一組
if (step == n + 1) {
// 判斷有沒有選到自己重複的
for (int i = 1; i <= n; i++) {
if (arr[i].gift.equals(arr[a[i]].gift))
return;
}
for (int i = 1; i <= n; i++) {
if (i != 1)
this.strBuf.append(",");
this.strBuf.append(arr[i].name + " " + arr[a[i]].gift);
}
strBuf.append("\n");
k++;
return;
}
for (int i = 1; i <= n; i++) {
if (b[i] == 0) {
a[step] = i;
b[i] = 1;
dfs(step + 1);
b[i] = 0;
}
}
return;
}
void ans() {
System.out.println(this.k);
System.out.print(this.strBuf);
}
}
class Human {
String name;
String gift;
}
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
P p = new P(n);
for (int i = 1; i <= n; i++) {
p.arr[i] = new Human();
p.arr[i].name = sc.next();
p.arr[i].gift = sc.next();
}
p.dfs(1);
p.ans();
}
}
```
### [C_MM46-易] 複數運算
#### C++:
```cpp=
#include <iostream>
using namespace std;
int main()
{
int n,a,b,c,d;
char optor;
cin >> n;
while(n-->0)
{
cin >> optor >> a >> b >> c >> d;
if (optor == '+')
cout << (a + c) << " " << (b + d) << endl;
else if (optor == '-')
cout << (a - c) << " " << (b - d) << endl;
else if (optor == '*')
cout << (a * c) - (b * d) << " " << (b * c) + (a * d) << endl;
else if (optor == '/')
cout << ((a * c) + (b * d)) / (c * c + d * d) << " " << ((b * c) - (a * d)) / (c * c + d * d) << endl;
}
return 0;
}
```
### [C_MM48-易] F91
#### C++:
```cpp=
#include <iostream>
using namespace std;
int f91(int n)
{
if(n<=100)
return f91(f91(n+11));
else if(n>=101)
return n-10;
}
int main()
{
int k,n;
cin >> k;
while(k-->0)
{
cin >>n;
cout << f91(n) << endl;
}
return 0;
}
```
### [C_MM49-易] 連續1的倍數
#### C++:
```cpp=
#include <iostream>
using namespace std;
int main()
{
int k, n, num=1, count = 1;
cin >> k;
while(k-->0)
{
cin >> n;
for(; (num %= n)!=0; count++)
{
num = 10*num+1;
}
cout << count << endl;
num = 1;
count = 1;
}
return 0;
}
```
### <font color="#f00">#[C_MM50-易] 分贓 </font>
#### C++:
```cpp=
#include <bits/stdc++.h>
#define _for(i,a,b) for(int i=a;i<b;i++)
using namespace std;
int readInt(){
int x;
cin >> x;
return x;
}
int main(){
int k=readInt();
int total=0,half;
int w[k];
_for(i,0,k)
total+=w[i]=readInt();
half=total/2;
int bag[half+1];
memset(bag,0,sizeof(bag));
_for(i,0,k)
for(int j=half;j>=w[i];j--)
bag[j]=max(bag[j-w[i]]+w[i],bag[j]);
cout << (int)abs(total-2*bag[half]) << endl;
return 0;
}
```
#### Java:
```java=
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int k = sc.nextInt();
int[] w = new int[k];
int total = 0;
for (int i = 0; i < k; i++)
total += w[i] = sc.nextInt();
int half = total / 2;
int[] bag = new int[half + 1];
for (int i = 0; i < k; i++)
for (int j = half; j >= w[i]; j--)
bag[j] = Math.max(bag[j - w[i]] + w[i], bag[j]);
System.out.println(Math.abs(total - 2 * bag[half]));
}
}
```
###### tags: `e tutor` `數學 I(Mathematics)`
> [name=風思] 當作練習C++~
> [color=#770aaa]
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