Uva 10790 How Many Points of Intersection?

How Many Points of Intersection?

We have two rows. There are a dots on the toprow and b dots on the bottom row. We draw line segments connecting every dot on the top row with every dot on the bottom row. The dots are arranged in such a way that the number of internal intersections among the line segments is maximized. To achieve this goal we must not allow more than two line segments to intersect in a point. The intersection points on the top row and the bottom are not included in our count; we can allow more than two line segments to intersect on those two rows. Given the value of a and b, your task is to compute P(a, b), the number of intersections in between the two rows. For example, in the following figure a = 2 and b = 3. This figure illustrates that P(2, 3) = 3.

Input 

Each line in the input will contain two positive integers a(0 < a20000) and b (0 < b20000).Input is terminated by a line where both a and b are zero. Thiscase should not be processed. You will need to process at most 1200 sets of inputs.

Output 

For each line of input, printin a line the serial of output followed by the value of P(a, b). Look atthe output for sample input for details. You can assume that the output for the test cases will fit in 64-bit signed integers.

Sample Input 

2 2
2 3
3 3
0 0

Sample Output 

Case 1: 1
Case 2: 3
Case 3: 9

题目大意：如果上面有a个点，下面有b个点，叫你计算交点的个数。

那么相交点的数目为：a*(a-1)*b*(b-1)/4;

这道题就是一道纯公式题目，数学如果学好来，然后公式推导出来，这道题目就AC来，注意题目要求说的要用long long int；

#include<stdio.h>
#include<math.h>
#include<iostream>
using namespace std;
int main() {
	long long a,b;
	long long sum;
	int t=1;
	while( cin >>a >>b ) {
		if(a==0 && b==0)
			break;
		sum =a*(a-1)*b*(b-1)/4; 
		cout<<"Case "<<t++<<": "<< sum <<endl;
	}

	return 0;
}