BZOJ3560:DZY Loves Math V 数学

思路:

首先我们需要知道欧拉函数的计算方法:

\[n=p_1^{q_1}p_2^{q_2}...p_m^{q_m}\rightarrow\phi(n)=n\frac{p_1-1}{p_1}\frac{p_2-1}{p_2}...\frac{p_m-1}{p_m}\]

\[\phi(n)=p_1^{q_1}\frac{p_1-1}{p_1}p_2^{q_2}\frac{p_2-1}{p_2}...p_m^{q_m}\frac{p_m-1}{p_m}\]

这证明我们可以将所有的质因数分开考虑.

我们对于所有质因数计算其贡献,然后乘到一起.(这东西维护一个前缀和就行了)

#include<cstdio>
#include<cstring>
#include<cctype>
#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;
 
#define N 10000010
int p[N/10],ins[N],cnt;bool notp[N];vector<int>v[N/10];
inline void pre(){
    register int i,j;
    for(i=2;i<=10000000;++i){
        if(!notp[i])p[++cnt]=i,ins[i]=cnt;
        for(j=1;j<=cnt&&i*p[j]<=10000000;++j){
            notp[i*p[j]]=1;
            if(i%p[j]==0)break;
        }
    }
}
 
static const int mod=(1e9)+7;
inline int ksm(int x,int y){
    int t=x,res=1;for(;y;y>>=1,t=(long long)t*t%mod)if(y&1)res=(long long)res*t%mod;return res;
}
inline int inv(int x){
    return ksm(x,mod-2);
}
inline void inc(int&x,int y){
    if((x+=y)>=mod)x-=mod;
}
 
int seq[1000010],num;vector<int>vv[1000010];
 
int pref[40];
 
int main(){ 
    pre();
     
    int n;scanf("%d",&n);register int i,j,k;
    int x,nowadd;
    while(n--){
        scanf("%d",&x);if(x==1)continue;
        for(i=1;i<=cnt&&p[i]*p[i]<=x;++i){
            nowadd=0;while(x%p[i]==0)++nowadd,x/=p[i];
            if(nowadd)v[i].push_back(nowadd);
        }if(x)v[ins[x]].push_back(1);
    }
     
    for(i=1;i<=cnt;++i)if((int)v[i].size()!=0){
        seq[++num]=p[i];
        for(j=0;j<(int)v[i].size();++j)vv[num].push_back(v[i][j]);
    }
     
    if(num==0)puts("1");
    else{
        int res=1,Mx,mi,tans;
        for(k=1;k<=num;++k){
            for(Mx=0,i=0;i<vv[k].size();++i)Mx=max(Mx,vv[k][i]);
            for(pref[0]=1,mi=seq[k],i=1;i<=Mx;++i,mi=(long long)mi*seq[k]%mod)inc(pref[i]=pref[i-1],mi);
            for(tans=1,i=0;i<vv[k].size();++i)tans=(long long)tans*pref[vv[k][i]]%mod;
            tans=(long long)(tans+mod-1)*(seq[k]-1)%mod;
            tans=(long long)tans*inv(seq[k])%mod;
            res=(long long)res*(tans+1)%mod;
        }
        printf("%d",res);
    }
     
    return 0;
}