#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;
}