输入两个整数，求这两个整数的和是多少。

#include <iostream>
using namespace std;
int main () {
    int a, b;
    cin >> a >> b;
    cout << a + b << endl;
    return 0;
}

差

#include<iostream>
using namespace std;
int main()
{
    int A,B,C,D;
    cin>>A>>B>>C>>D;
    cout<<"DIFERENCA = "<< A * B - C * D;
    return 0;
}

圆的面积

#include <iostream>
#include <cstdio> // 运用scanf/printf函数
#define PI 3.14159

using namespace std;

int main()
{
    double R, A;
    scanf("%lf", &R);
    A = R * R * PI;
    printf("A=%.4lf", A);
    return 0;
}

平均数1

#include<iostream>
using namespace std;
int main()
{
    double a,b;
    cin>>a>>b;
    printf("MEDIA = %.5lf",(a*3.5+b*7.5)/11);//11=3.5+7.5
    return 0;
}

工资

#include <cstdio> 
using namespace std;
int main()
{
    int number,hour;
    double money;
    scanf("%d%d%lf " , &number, &hour, &money);
    printf("NUMBER = %d\n ", number);
    printf("SALARY = U$ %.2lf\n ", hour * money);
    return 0;
}

油耗

#include<iostream>
using namespace std;
int main()
{
    int X;
    double Y;
    cin>>X>>Y;
    printf("%.3lf km/l",X/Y);
    return 0;
}

两点间的距离

#include <iostream>
#include <cmath>
using namespace std;
int main()
{
    double x1, y1, x2, y2;
    cin >> x1 >> y1 >> x2 >> y2;
    printf("%.4lf", sqrt(pow(x2 - x1, 2) + pow(y2 - y1, 2)));
    return 0;
}

钞票

#include <iostream>
using namespace std;
int main()
{
    int n, a[7] = {100, 50, 20, 10, 5, 2, 1};
    cin >> n;
    printf("%d\n", n);
    for (int i = 0; i < 7; i ++ )
    {
        printf("%d nota(s) de R$ %.2lf\n", n / a[i], a[i]);
        n %= a[i];
    }
    return 0;
}

时间转化

#include <iostream>

using namespace std;

int main()
{
    int t;
    cin >> t;
    int h = t / 3600;
    int m = t % 3600 / 60;
    int s = t % 60;
    cout << h << ':' << m << ':' << s << endl;
    return 0;
}

简单计算

#include<iostream>
using namespace std;
int main()
{
    int code,n;
    double price,sum=0.0;
    for(int i=1;i<=2;i++)
        cin>>code>>n>>price,sum+=n*price;
    printf("VALOR A PAGAR: R$ %.2lf",sum);
    return 0;
}

燃料消耗

• 这个题数据范围挺大，小心溢出

#include<iostream>
using namespace std;
int a, b;
int main() {
    scanf("%d%d", &a, &b);
    printf("%.3lf", a * 1.000 / 12 * b); //先除再乘，以免double溢出
    return 0;
}

钞票和硬币

• 数值放大 解法和上面类似

#include <iostream>
#include <cstdio>
using namespace std;
int main()
{
    double n;
    cin >> n;
    int m = (int)(n * 100);
    double a[12] = {10000, 5000, 2000, 1000, 500, 200, 100, 50, 25, 10, 5, 1};
    int ans[12] = {0};
    for (int i = 0; i < 12; i++)
    {
        int cnt = 0;
        while (m >= a[i])
        {
            m -= a[i];
            cnt++;
        }
        ans[i] = cnt;
    }
    puts("NOTAS:");
    for (int i = 0; i < 6; i++)
        printf("%d nota(s) de R$ %.2lf\n", ans[i], (double)a[i] / 100);
    puts("MOEDAS:");
    for (int i = 6; i < 12; i++)
        printf("%d moeda(s) de R$ %.2lf\n", ans[i], (double)a[i] / 100);
    return 0;
}