题目大意:求上面的点和下面的点的连线的交点有多少个
解题思路:纯数学题,推出公式a*(a-1)*b*(b-1)/4
#include<cstdio>
int main() {
long long num1, num2;
int mark = 1;
while(scanf("%lld %lld", &num1, &num2) != EOF){
if(num1 == 0 && num2 == 0)
break;
printf("Case %d: ", mark++);
long long count = 0;
printf("%lld\n",num1*(num1 - 1)*num2*(num2 - 1)/4);
}
return 0;
}