(组合)Binomial Showdown

最新推荐文章于 2019-03-18 20:48:45 发布

转载最新推荐文章于 2019-03-18 20:48:45 发布 · 80 阅读

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原文链接：http://www.cnblogs.com/ilmxt/archive/2012/07/06/2578691.html

本文介绍了一种计算组合数的方法，即从n个不同元素中选择k个元素的方式数量，不考虑顺序。通过使用对数和指数运算来避免大整数运算过程中可能产生的溢出问题。

求组合数

Description

In how many ways can you choose k elements out of n elements, not taking order into account?
Write a program to compute this number.

Input

The input will contain one or more test cases.
Each test case consists of one line containing two integers n (n>=1) and k (0<=k<=n).
Input is terminated by two zeroes for n and k.

Output

For each test case, print one line containing the required number. This number will always fit into an integer, i.e. it will be less than 2³¹.
Warning: Don't underestimate the problem. The result will fit into an integer - but if all intermediate results arising during the computation will also fit into an integer depends on your algorithm. The test cases will go to the limit.

Sample Input

Sample Output

6
252
13983816

Author

Ulm Local 1997

//Binomial Showdown
#include<stdio.h>
#include<math.h>
double f(double n,double m)
{
     double i,s1,s2;
     if(n==m) return(1);
          if(m<n/2.0)
          {
               m=n-m;
          }
          s1=0;s2=0;
          for(i=m+1;i<=n;i++)
               s1+=log(i);
          for(i=2;i<=n-m;i++)
               s2+=log(i);
          return(exp(s1-s2));
}
int main()
{
     double i,n,m;
     while(scanf("%lf%lf",&n,&m)!=EOF)
     {
          if(n==0&&m==0) break;
          if(n==1&&m==0) printf("1\n");
          else
          {
           printf("%.0f\n",f(n,m));
          }
     }
     return(0);
}

转载于:https://www.cnblogs.com/ilmxt/archive/2012/07/06/2578691.html