POJ_1942_Paths on a Grid 组合数学

跟小伙伴们出去吃饭。

题意：

一个(n+1)*(m+1)的网格，从左下的顶点出发，只能向上或向右走，要走到右上顶点，问有多少种走法？

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.
首先，肯定要走n+m步，其中必然有n部向上走m步向右走，所以答案就是C(N+M,N)，n和m都比较大，可以使用第三种求的方法。
C(n,m)=m!/(n!*(n-m)!),上下同除(n-m)!就可以在O(n)时间中求出来答案了，然后题目中没有n和m都到10^9级别的数据。。。就过了。。。不过这种求组合数的方法还是值得一记的。

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
using namespace std;
long long n,m;
int main(){
while(scanf("%I64d%I64d",&n,&m)!=EOF){
if(!n&&!m)	break;
long long sum=n+m;
long long minn=min(n,m);
long long ans=1.0;
for(long long i=1;i<=minn;++i)
ans=ans*(sum-minn+i)/i;
printf("%I64d\n",ans);
}
return 0;
}