poj942Paths on a Grid(组合数学)

最新推荐文章于 2020-04-05 17:11:32 发布

原创最新推荐文章于 2020-04-05 17:11:32 发布 · 460 阅读

0 ·

CC 4.0 BY-SA版权

数论专栏收录该内容

15 篇文章

订阅专栏

本文介绍了一个数学问题：从网格的左下角到右上角，只允许向上或向右移动，有多少种不同的路径？通过组合数学的方法求解，并提供了一个C++实现。

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 24411		Accepted: 6050

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(杰作), isn't it? Repeating the procedure(程序) 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(投入) contains several testcases. Each is specified(指定) by two unsigned 32-bit integers(整数) n and m, denoting(表示) the size of the rectangle. As you can observe, the number of lines of the corresponding grid is one more in each dimension(维). Input is terminated(终止) by n=m=0.

Output

For each test case output(输出) on a line the number of different art works that can be generated(形成) 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(承担) that this number fits into a 32-bit unsigned integer.

Sample Input

5 4
1 1
0 0

Sample Output

126
2
参考了我大姐的博客   吊的一匹  长姿势了～～～
#include <iostream>
#include<stdio.h>
#define min(a,b) a>b?b:a
using namespace std;
unsigned cm( unsigned n,unsigned m)
  {  double cnt=1.0;
      while(m>0)
        cnt*=(double)(n--)/(double)(m--);
        cnt+=0.5;
      return (unsigned)cnt;
  }
int main()
{
     unsigned a,b;
    while(~scanf("%u%u",&a,&b))
    {

        if(!a&&!b)
            break;
        unsigned n=a+b;
        unsigned m=min(a,b);
        printf("%u\n",cm(n,m));
    }


    return 0;
}