UVA

UVA - 10375

Choose and divide

 

Choose and divide

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4053		Accepted: 1318

Description

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).

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

Source

Waterloo local 1999.10.02

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唯一分解定理

太诡异了，uva AC

poj一直TLE，连白书的标解都WA

//
//  main.cpp
//  poj2613
//
//  Created by Candy on 10/20/16.
//  Copyright ? 2016 Candy. All rights reserved.
//

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
const int N=1e4+5;
int p,q,r,s;
int prime[N],cnt=0,e[N],vis[N];
void era(int n){
    int m=sqrt(n)+1;
    for(int i=2;i<=m;i++) if(!vis[i])
        for(int j=i*i;j<=n;j+=i) vis[j]=1;
    for(int i=2;i<=n;i++) if(!vis[i]) prime[++cnt]=i;
}
inline void mul(int x,int d){//x^d
    for(int i=1;i<=cnt&&x!=1;i++)
        while(x%prime[i]==0){
            x/=prime[i];
            e[i]+=d;
        }
}
void fac(int x,int d){//printf("%d %d
",x,d);
    for(int i=1;i<=x;i++) mul(i,d);
}
inline int fastPow(int a,int b){
    int ans=1;
    for(;b;b>>=1,a*=a)
        if(b&1) ans*=a;
    return ans;
}
int main(int argc, const char * argv[]){
    era(10000);//cout<<"p";
    while(scanf("%d%d%d%d",&p,&q,&r,&s)!=EOF){
        memset(e,0,sizeof(e));
        fac(p,1);
        fac(q,-1);
        fac(p-q,-1);
        fac(r,-1);
        fac(s,1);
        fac(r-s,1);
        double ans=1;
        for(int i=1;i<=cnt;i++){
            if(e[i]>0) ans*=(double)fastPow(prime[i],e[i]);
            else if(e[i]<0) ans/=(double)fastPow(prime[i],-e[i]);
//            ans*=pow(prime[i],e[i]);
        }
        printf("%.5f
",ans);
    }
 
    return 0;
}