poj 1942 Paths on a Grid 水组合

题意：

给定一个矩形网格的长m和高n，其中m和n都是unsigned int32类型，一格代表一个单位，就是一步，求从左下角到右上角有多少种走法，每步只能向上或者向右走

显然答案=C(n+m,m)=C(n+m,n)。

一共走n+m步，在其中选m步向右走即可。

Paths on a Grid

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 24192 Accepted: 5995

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 41 10 0

Sample Output

1262

Source

Ulm Local 2002

#include<cstdio>#include<string>#include<cstring>#include<iostream>#include<cmath>#include<algorithm>#include<vector>using namespace std;#define all(x) (x).begin(), (x).end()#define for0(a, n) for (int (a) = 0; (a) < (n); (a)++)#define for1(a, n) for (int (a) = 1; (a) <= (n); (a)++)typedef long long ll;typedef pair<int, int> pii;const int INF =0x3f3f3f3f;ll C(ll n,ll m){   ll ans=1;   for(ll i=1;i<=m;i++)   {       ans=ans*(n-i+1)/i;   }   return ans;}int main(){    ll n,m;    while(~scanf("%lld%lld",&n,&m)&&(n||m))    {       if(m>n) swap(m,n);       n=m+n;       printf("%lld\n",C(n,m));    }   return 0;}