UVA 10790 (13.08.06)

 How Many Points of Intersection? 

We have two rows. There are a dots on the toprow andb 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 computeP(a, b), the number of intersections in between the two rows. For example, in the following figurea = 2 and b = 3. This figure illustrates thatP(2, 3) = 3.

Input

Each line in the input will contain two positive integers a(0 <a20000) andb (0 < b20000).Input is terminated by a line where botha 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 ofP(a, b). Look atthe output for sample input for details. You can assume that the output for the test cases will fit in64-bit signed integers.

Sample Input

2 2
2 3
3 3
0 0

Sample Output

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

解题思路请参考: http://www.cnblogs.com/xiaocai905767378/archive/2011/04/27/2030213.html

注意啊注意: 用long long

row_a = 20000

roe_b = 20000 时, 会超啊~

AC代码:

#include<stdio.h>

long long a, b;
int cas = 0;

int main() {
    while(scanf("%lld %lld", &a, &b) != EOF) {
        if(a == 0 && b == 0)
            break;
        long long ans = (a * (a-1) * b * (b-1) / 4);
        printf("Case %d: %lld\n", ++cas, ans);
    }
    return 0;
}