How Many Points of Intersection? 

a dots on the top row 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 

a ( 0 < a20000) and b ( 0 < b20000). Input is terminated by a line where both a and b

Output 

P(a, b). Look at the 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 22 3 3 3 0 0


Sample Output 


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


Problem setter: Md. Bahlul Haider

Special Thanks: Shahriar Manzoor

Miguel Revilla 2004-12-10





    题意：给出两个数字n，m分别代表上和下有多少个点，求这些点连成的线会有多少个交点。输出。。。找到之间的关系就可以根据式子求出来。




#include<stdio.h>
#include<string.h>
#include<stdlib.h>

int main()
{
    long long int n,m;
    int k = 0;
    while(scanf("%lld%lld",&n,&m)!=EOF)
    {
        k++;
        if(n == 0 && m == 0)
        {
            break;
        }
        long long int sum = 0;
        sum = (n-1)*n/2 * (m-1)*m/2;
        printf("Case %d: %lld\n",k,sum);
    }
    return 0;
}