文章标题 POJ 1942： Paths on a Grid（数学）

原创于 2016-11-16 17:30:50 发布 · 389 阅读

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本文介绍了一个经典的组合数学问题——在一个n*m的网格中，从右下角走到左上角，仅允许向上或向右移动，计算可行路径的数量。文章提供了详细的解析方法和高效算法实现。

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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=a 2+2ab+b 2). 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

题意：在一个n*m的方格中从右下角走到左上角，只能往上或者往右走。问有多少种方法。
分析：从右下角走到左上角，所以得往右走m步，往上走n步，所以问题就转化成从m+n中取n步的数目。但当n过大的时候，不能直接求n和m的阶乘，所以可以用拆分阶乘，逐项相除，再乘以前面所有项之积的方法。
代码：

#include<iostream>
#include<string>
#include<cstdio>
#include<cstring>
#include<vector>
#include<math.h>
#include<map>
#include<queue> 
#include<algorithm>
using namespace std;
const int inf = 0x3f3f3f3f;
unsigned n,m;
unsigned Ans(unsigned n,unsigned m){
    unsigned com=n+m;
//两个想加
    unsigned b=min(n,m);
//取两者较小的，可以减少计算
    double ans=1.0;
    while (b>0){
        ans*=(double)(com--)/(b--);
//分子的一项除以分母的一项，注意得类型转换成double型
    }
    ans+=0.5;
//加0.5，用于四舍五入
    return (unsigned)ans;
//返回时也得类型转换
}
int main ()
{
    while (cin>>n>>m){
        if (n==0&&m==0) break;
        cout<<Ans(n,m)<<endl; 
    }
    return 0;
}