UVA 10375 Choose and divide

最新推荐文章于 2020-02-24 15:05:28 发布

「已注销」

最新推荐文章于 2020-02-24 15:05:28 发布

阅读量268

点赞数

CC 4.0 BY-SA版权

分类专栏：数学

本文链接：https://blog.youkuaiyun.com/Rlt1296/article/details/52670874

数学专栏收录该内容

54 篇文章

订阅专栏

本文介绍了一种通过唯一分解定理来高效计算两个组合数比值的方法。利用质数及其指数进行运算，避免了直接计算阶乘所带来的巨大数值计算问题。通过记录每次除法操作中质数因子的变化情况，最终求得精确结果。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

The binomial coefficient C(m, n) is defined as
C(m, n) = m!
(m − n)! n!
Given four natural numbers p, q, r, and s, compute the the result of dividing C(p, q) by C(r, s).
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.
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
Sample Output
0.12587
505606.46055
1.28223
0.48996
2.00000
3.99960

用唯一分解定理，记录次数，最后求出答案。

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;
const int N=10005;
int p,q,r,s,cnt,pri[N],c[N];
bool vis[N];
double ans;
void add(int x,int y)
{
    for(int i=1;i<=x;i++)
    {
        int tmp=i;
        for(int j=1;j<=cnt&&y!=0;j++)
            while(tmp%pri[j]==0)
            {
                tmp/=pri[j];
                c[j]+=y;
            }
    }
}
void euler()
{
    vis[1]=1;
    for(int i=2;i<=10000;i++)
    {
        if(!vis[i])
            pri[++cnt]=i;
        for(int j=1;j<=cnt&&pri[j]*i<=10000;j++)
        {
            vis[i*pri[j]]=1;
            if(i%pri[j]==0)
                break;
        }
    }
}
int main()
{
    euler();
    while(scanf("%d%d%d%d",&p,&q,&r,&s)==4)
    {
        ans=1;
        memset(c,0,sizeof(c));
        add(p,1);
        add(p-q,-1);
        add(q,-1);
        add(r,-1);
        add(r-s,1);
        add(s,1);
        for(int i=1;i<=cnt;i++)
            ans*=pow(pri[i],c[i]);
        printf("%.5f\n",ans);
    }
    return 0;
}