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=;

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

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

 int e[maxn];

 int power (int a,int b){
     int ans=;
     while (b){
         if (b&)
             ans*=a;
         a*=a;
         b>>=;
     }
     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,,sizeof e);
         int ma=max (p,r);
         for (int i=;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=;
         for (int i=;i<tot;i++){
             if (e[i]>=)
                 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 ;
 }