POJ：1942 Paths on a Grid（组合）

Paths on a Grid

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 24968 Accepted: 6215

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

题目大意：给你一个m*n的表格，让你从左下方走到右上方，问总共有多少种走法。

解题思路：从左下方走到右上方，一定走了m+n步，选择其中n步（或m步）放到计划中固定，代表在哪一点向上走（向右走），那么就是求C(n+m,n或m)，为了减少时间，选择较小的作为第二项。

代码如下：

#include <cstdio>#include <cmath> double jiecheng(double n,double m)//求cnm {double jieguo=1.0;while(m){jieguo=jieguo*((n--)/(m--));}return jieguo;}int main(){double n,m,ans;while(scanf("%lf%lf",&n,&m)!=EOF){if(n==0&&m==0)//题目要求 {break;}if(n>m){ans=jiecheng(n+m,m);}else{ans=jiecheng(n+m,n);}__int64 shuchu=(__int64)round(ans);//round是四舍五入函数，用int64的类型，int爆表了 printf("%I64d\n",shuchu);}return 0;}