Paths on a Grid POJ 1942

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=a 2+2ab+b 2). 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：给你一个几个格子的长方形，让你从左下角走到右下角有多少种走法；

思路：这道题其实是

向上走有n中方法，向右走有m种方法，先把就相当于n+m个位置选n个或选m个（选了n个那m个就固定了，选m个同理剩下n个也固定了）。所以代码主要是如何求组合数

#include<stdio.h>int main(){unsigned long long a,b,e,f,i,j,ans;while(scanf("%lld%lld",&a,&b)!=EOF){if(a==0&&b==0)break;else{ans=1;e=a;if(e<b)e=b;f=a+b;for (i=e+1,j=1;i<=f;i++,j++)    {        ans=ans*i/j;     } printf("%lld\n",ans);}}return 0;}