POJ 1942 Paths on a Grid (水题)

Description

Imagine you are attending your math lesson at school. Once again, you are bored because your teacher tells things that you already mastered years ago (this time he's explaining that (a+b)²=a²+2ab+b²). So you decide to waste your time with drawing modern art instead.

Fortunately you have a piece of squared paper and you choose a rectangle of size n*m on the paper. Let's call this rectangle together with the lines it contains a grid. Starting at the lower left corner of the grid, you move your pencil to the upper right corner, taking care that it stays on the lines and moves only to the right or up. The result is shown on the left:

Really a masterpiece, isn't it? Repeating the procedure one more time, you arrive with the picture shown on the right. Now you wonder: how many different works of art can you produce?

Input

The input contains several testcases. Each is specified by two unsigned 32-bit integers n and m, denoting the size of the rectangle. As you can observe, the number of lines of the corresponding grid is one more in each dimension. Input is terminated by n=m=0.

Output

For each test case output on a line the number of different art works that can be generated using the procedure described above. That is, how many paths are there on a grid where each step of the path consists of moving one unit to the right or one unit up? You may safely assume that this number fits into a 32-bit unsigned integer.

Sample Input

5 4
1 1
0 0

Sample Output

126
2

题意很清晰，就是算出从一个对角线到另一个对角线有多小走法（只能向上，向右走）。
分析：一个矩阵，它有行有列，要到达对角线，必定有通过所有的行和列，那么存在两种情况
         当行(n)==列(m)，即刚好走完列和行，在m+n个里选m或n个组合得C（m+n，m）或C(m+n,n)。
         当两者不相等，就要看那个先到，所有最小的必定先到达。

#include <iostream>
using namespace std;
int main()
{
    long long n,m,i;
    while(cin>>n>>m&&(n+m!=0))
    {
        long long sum=1;
        long long a=n+m;
        long long b=n<m?n:m;
        for(i=1;i<=b;i++)
          sum=sum*(a--)/i;
        cout<<sum<<endl;
    }
}