nyoj（简单数学）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).

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

来源

原创

上传者

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

<script type="text/java script">BAIDU_CLB_fillSlot("771048");

点击复制链接 与好友分享! 回本站首页

<script> function copyToClipBoard(){ var clipBoardContent=document.title + '\r\n' + document.location; clipBoardContent+='\r\n'; window.clipboardData.setData("Text",clipBoardContent); alert("恭喜您！复制成功"); }

<script>window._bd_share_config={"common":{"bdSnsKey":{},"bdText":"","bdMini":"2","bdMiniList":false,"bdPic":"","bdStyle":"0","bdSize":"24"},"share":{}};with(document)0[(getElementsByTagName('head')[0]||body).appendChild(createElement('script')).src='http://bdimg.share.baidu.com/static/api/js/share.js?v=89860593.js?cdnversion='+~(-new Date()/36e5)];