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;
}