POJ1942——Paths on a Grid(组合数学)

Paths on a Grid

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

题目大意：

　　　　给定一个M*N的方格。问有多少种走法使其从左下角到右上角。

解题思路：

　　　　简单的组合数学。

　　　　从左下角到右上角。毕竟要向右M步，向上N步。共计M+N步。求Com[M+N][M]即可。

　　　　PS：Com[M+N][M]=Com[M+N][N] 在求Com的时候，可以选择min(M,N)来进行计算。否则超时。。。

　　　　PS2：注意被调写法的正确性。若先算分子后算分母会爆longlong。

Code：

/*************************************************************************
    > File Name: poj1942.cpp
    > Author: Enumz
    > Mail: 369372123@qq.com
    > Created Time: 2014年10月21日 星期二 20时12分35秒
 ************************************************************************/

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<string>
#include<cstring>
#include<list>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<algorithm>
#define MAXN 100000
using namespace std;
long long c(long long a,long long b)
{
    long long ret=1;
    for (long long i=1;i<=a;i++)
    {
        ret=ret*(b--)/i;   /*注意其正确性，每经过i个数，必有一个能被i整除*/
    }
    return ret;
}
int main()
{
    long long a,b;
    cout<<c(2,3)<<endl;
    while (cin>>a>>b)
    {
        if (a>b) swap(a,b);
        if (!a&&!b) break;
        cout<<c(a,a+b)<<endl;
    }
    return 0;
}