UVA 10375 Choose and divide

n! 分解素因子　　快速幂

 

ei=[N/pi^1]+ [N/pi^2]+ …… + [N/pi^n]  其中[]为取整

ei 为数 N！中pi 因子的个数；

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;

const int maxn=10005;

int sign[maxn];
int pri[maxn];
int tot;

void getpri (){
    memset (sign,0,sizeof sign);
    sign[0]=sign[1]=1;
    for (int i=2;i<sqrt (maxn+0.5);i++)
        if (!sign[i])
            for (int j=i*i;j<maxn;j+=i)
                sign[j]=1;
    tot=0;
    for (int i=2;i<maxn;i++)
        if (!sign[i])
            pri[tot++]=i;
}

int e[maxn];

int power (int a,int b){
    int ans=1;
    while (b){
        if (b&1)
            ans*=a;
        a*=a;
        b>>=1;
    }
    return ans;
}

int main (){
    getpri ();//cout<<tot<<endl;
    int p,q,r,s;
    while (~scanf ("%d%d%d%d",&p,&q,&r,&s)/*cin>>p>>q>>r>>s*/){
        memset (e,0,sizeof e);
        int ma=max (p,r);
        for (int i=0;i<tot;i++){
            int temp=pri[i];
            while (temp<=ma){
                e[i]+=p/temp+s/temp+(r-s)/temp;//if (i==0) cout<<ma<<" ";
                e[i]-=r/temp+q/temp+(p-q)/temp;
                temp*=pri[i];
            }
        }
        double ans=1;
        for (int i=0;i<tot;i++){
            if (e[i]>=0)
                ans*=1.0*power (pri[i],e[i]);
            else ans/=1.0*power (pri[i],-e[i]);//cout<<e[i]<<":"<<pri[i]<<"=";//<<ans<<" ";
        }
        printf ("%.5f\n",ans);
    }
    return 0;
}