UVA10375 Choose and divide 质因数分解

质因数分解：

Time Limit: 3000MS

Memory Limit: Unknown

64bit IO Format: %lld & %llu

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Description

Problem D: Choose and divide

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

Source

Root :: AOAPC II: Beginning Algorithm Contests (Second Edition) (Rujia Liu) :: Chapter 10. Maths :: Examples

Root :: AOAPC I: Beginning Algorithm Contests (Rujia Liu) :: Volume 6. Mathematical Concepts and Methods

Root :: Competitive Programming 2: This increases the lower bound of Programming Contests. Again (Steven & Felix Halim) :: Mathematics :: Combinatorics ::

option=com_onlinejudge&Itemid=8&category=405" style="color:blue; text-decoration:none">Binomial
Coefficients

Root :: Competitive Programming 3: The New Lower Bound of Programming Contests (Steven & Felix Halim) :: Mathematics :: Combinatorics :: Binomial
Coefficients

Root :: Competitive Programming: Increasing the Lower Bound of Programming Contests (Steven & Felix Halim) :: Chapter 5. Mathematics :: Combinatorics

Root :: Prominent Problemsetters :: Gordon V. Cormack

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#include <iostream>

#include <cstring>

#include <cstdio>

#include <algorithm>

using namespace std;

const int maxn=10010;

int p,q,r,s;

int prime[maxn],pn;

long long int fnum[maxn],pnum[maxn];

bool vis[maxn];

void pre_init()

{

    memset(vis,true,sizeof(vis));

    for(int i=2; i<maxn; i++)

    {

        if(i%2==0&&i!=2) continue;

        if(vis[i]==true)

        {

            prime[pn++]=i;

            for(int j=2*i; j<maxn; j+=i)

            {

                vis[j]=false;

            }

        }

    }

}

void fenjie_x(int x,long long int* arr)

{

    for(int i=0; i<pn&&x!=1; i++)

    {

        while(x%prime[i]==0)

        {

            arr[i]++;

            x/=prime[i];

        }

    }

}

void fenjie(int x,long long int* arr)

{

    for(int i=2; i<=x; i++)

        fenjie_x(i,arr);

}

void jianshao()

{

    for(int i=0; i<pn; i++)

    {

        long long int Min=min(fnum[i],pnum[i]);

        fnum[i]-=Min;

        pnum[i]-=Min;

    }

}

int main()

{

    pre_init();

    while(scanf("%d%d%d%d",&p,&q,&r,&s)!=EOF)

    {

        memset(pnum,0,sizeof(pnum));

        memset(fnum,0,sizeof(fnum));

        fenjie(p,pnum);fenjie(s,pnum);fenjie(r-s,pnum);

        fenjie(q,fnum);fenjie(r,fnum);fenjie(p-q,fnum);

        jianshao();

        double ans=1.;

        for(int i=0; i<pn; i++)

        {

            while(pnum[i]--)

            {

                ans*=1.*prime[i];

            }

            while(fnum[i]--)

            {

                ans/=1.*prime[i];

            }

        }

        printf("%.5lf\n",ans);

    }

    return 0;

}