UVa 10790 - How Many Points of Intersection?

传送门UVa 10790 - How Many Points of Intersection?

题目挺简单, 观察一下就可以得出规律num = n * (n - 1) * m * (m - 1) / 4;

题目要求用long long输出... 于是我就声明了一个变量

long long sum

然后用

int m, n;

接收.

可是这样一来就错了...

后来求助了失踪同学, 了解到如果用int类型的m, n, 在计算n * (n - 1) * m * (m - 1) / 4就会溢出了.

涨姿势了..

#include <cstdio>#include <cmath>using namespace std;int main(){    //freopen("input.txt", "r", stdin);    long long m, n;    long long sum;    int count = 1;    int i, j;    while (scanf("%lld%lld", &m, &n))    {        sum = 0;        if (m == 0 && n == 0)            break;        sum = n * (n - 1) * m * (m - 1) / 4;        printf("Case %d: %lld\n", count++, sum);    }    return 0;}