POJ 1942Paths on a Grid（组合数学）

最新推荐文章于 2018-11-08 16:18:58 发布

原创最新推荐文章于 2018-11-08 16:18:58 发布 · 320 阅读

0 ·

CC 4.0 BY-SA版权

POJ初级专栏收录该内容

58 篇文章

订阅专栏

本文介绍了一种在数学课上通过从左下角走到右上角的网格路径绘制现代艺术的方法，并提供了一个算法来计算不同路径的数量。输入为网格大小，输出为可能的艺术作品数量。

Paths on a Grid

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 25192		Accepted: 6289

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

Source

从左下角到右上角所走步数。

#include <iostream>

using namespace std;

unsigned long long sum(unsigned long long n, unsigned long long m)
{
    unsigned long long s = 1, i, j;
    for(i = m+1, j = 1; i <= n; i++, j++)
    {
        s = s * i / j;
    }
    return s;
}
int main()
{
    unsigned long long n, m;
    while(cin>>n>>m)
    {
        unsigned long long x;
        if(m==0&&n==0)
        {
            return 0;
        }
        if(m > n)
        x = m;
        else
        x = n;
        cout<<sum(n+m, x)<<endl;
    }
}