nyoj（簡單數學）Oh, my Paper!

nyoj（簡單數學）Oh, my Paper!

Oh, my Paper!

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).

輸入

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

輸出

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

樣例輸入

2 3
1 1
0 0

樣例輸出

10
2

提示

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

來源

原創

上傳者

ACM_賀榮偉

思路：（直接用偉哥的）給你一張紙，n(行)*m（列）你要計算的是
算出從一個對角線到另一個對角線有多少走法（只能向上，向右走）。
分析：一個矩陣，它有行有列，要到達對角線，必定有通過所有的行和列，那麼存在兩種情況，當行(n)==列(m)，即剛好走完列和行，在m+n個裡選m或n個組合得C（m+n，m）或C(m+n,n)。
當兩者不相等，就要看那個先到，所有最小的必定先到達。

#include
#include
#include
#include
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=n0)
            sum*=(a--/(double)(b--));
        printf("%.lf\n",sum);
    }
}