唯一分解定理与组合数

Choose and divide
Description
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.

Samples
Input 复制
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
0.12587
505606.46055
1.28223
0.48996
2.00000
3.99960

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=1e5+7;
int num[maxn],prime[maxn],e[maxn];
int cou;
void oula()
{
	cou=0;	
	for(int i=2;i<=maxn;i++)
	{
		if(!num[i])prime[cou++]=i;
		for(int j=0;j<cou;j++)
		{
			if(i*prime[j]>maxn)break;
			num[i*prime[j]]=1;
			if(i%prime[j]==0)
			break;
		}
	}
}
void solve(ll n,ll d)
{
	for(ll i=0;i<cou;i++)
	{
		while(n%prime[i]==0)
		{
			n/=prime[i];
			e[i]+=d;
		}
		if(n==1)return ;
	}
}
ll q,r,p,s,s1,s2;
int main()
{
	memset(num,0,sizeof num);
	oula();
	while(cin>>p>>q>>r>>s)
	{
		double ans;
		ans=1;
		memset(e,0,sizeof e);
		for(int i=p;i>=1;i--) solve(i,1);
        for(int i=p-q;i>=1;i--) solve(i,-1);
        for(int i=q;i>=1;i--) solve(i,-1);
        for(int i=r-s;i>=1;i--) solve(i,1);
        for(int i=s;i>=1;i--) solve(i,1);
        for(int i=r;i>=1;i--) solve(i,-1);
        for(int i=0;i<cou;i++)
        {
        	if(e[i])ans*=pow(prime[i],e[i]);
		}
		printf("%.5lf\n",ans);
}
 }