nyoj（简单数学）Oh, my Paper!

标签：数学   算法   nyoj   

描述

Give you a piece of paper, n (row) *m (column) to calculate your is
Calculated from a diagonal line to another diagonal how many walk method (only upward or downward, left, and right away).

输入

The input contains several testcases. Each is specified by two unsigned 32-bit integers n and m The last test case is followed by a line that contains two zeroes. This line must not be processe

输出

For each test case output on a line that how many paths you can calculate

样例输入

2 3
1 1
0 0

样例输出

10
2

提示

the num is a little big ,you‘d better use unsigned

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原创

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ACM_贺荣伟

思路：（直接用伟哥的）给你一张纸，n(行)*m（列）你要计算的是
算出从一个对角线到另一个对角线有多少走法（只能向上，向右走）。
分析：一个矩阵，它有行有列，要到达对角线，必定有通过所有的行和列，那么存在两种情况，当行(n)==列(m)，即刚好走完列和行，在m+n个里选m或n个组合得C（m+n，m）或C(m+n,n)。
当两者不相等，就要看那个先到，所有最小的必定先到达。

#include<iostream>
#include<string.h>
#include<algorithm>
#include<stdio.h>
using namespace std;
int main()
{
    long long n,m;
    while(~scanf("%lld%lld",&n,&m),n||m)
    {
        long long a,b;
        a=n+m;
        b=n<m?n:m;
        double sum=1;
        while(b>0)
            sum*=(a--/(double)(b--));
        printf("%.lf\n",sum);
    }
}

nyoj（简单数学）Oh, my Paper!

标签：数学   算法   nyoj   

原文地址：http://blog.csdn.net/u012349696/article/details/41871389