nyoj（简单数学）Oh, my Paper!

Oh, my Paper!

时间限制：1000 ms  |  内存限制：65535 KB

难度：2

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 31 10 0

样例输出

102

提示

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

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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);    }}