Choose and divide-优快云博客

本文链接：https://blog.youkuaiyun.com/Rechard_chen/article/details/40747365

本文介绍了一种计算两个组合数比值的方法，并提供了一个高效的C++实现方案。该方案通过简化组合数公式来减少计算复杂度，适用于处理大规模数值。

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The binomial coefficient C(m,n) is defined as

         m!
C(m,n) = --------
         n!(m-n)!

Given four natural numbers p, q, r, and s, compute the the result of dividing C(p,q) by C(r,s).

The Input

Input consists of a sequence of lines. Each line contains four non-negative integer numbers giving values for p, q, r, and s, respectively, separated by a single space. All the numbers will be smaller than 10,000 with p>=q and r>=s.

The Output

For each line of input, print a single line containing a real number with 5 digits of precision in the fraction, giving the number as described above. You may assume the result is not greater than 100,000,000.

Sample Input

10 5 14 9
93 45 84 59
145 95 143 92
995 487 996 488
2000 1000 1999 999
9998 4999 9996 4998

Output for Sample Input

0.12587
505606.46055
1.28223
0.48996
2.00000
3.99960

题意：题目所求为组合 C(p,q) / C(r,s);那么可以乘以一个在除以一个；即将

         m!
C(m,n) = -------
         n!(m-n)!

化简成为 m * （m-1) * (m-2) . . . . . (m- n +1)/n!;    这个分子与分母的个数是一样的；

那么便有一下标程；

# include <iostream>
# include <cstdio>
using namespace std;

int main()
{
    //freopen("a.txt","r",stdin);
    int p,q,r,s;
    while(scanf("%d%d%d%d",&p,&q,&r,&s)!=EOF)
    {
        int i,n;
        double ans=1.0;
        //if (p - q < q)     q = p - q;
        //if (r - s < s)     s = r - s;
        for(i=1;i<=q||i<=s;i++)
        {
            if(i<=q)    ans=ans*(p-q+i)/i;
            if(i<=s)    ans=ans/(r-s+i)*i;
        }
        printf("%.5lf\n",ans);
    }
    return 0;
}