HDU 5525(Product-费马小定理)

给你一个数 N ,求其所有约数的积。

令 N=pa11∗pa22∗...∗pann

于是答案= ∏ni=1pi(ai+1)∗ai2∗∏nj=1(aj+1)ai+1(mod109+7)

= ∏ni=1p(ai+1)∗ai2∗∏nj=1(aj+1)ai+1(modϕ(109+7))i(mod109+7)
= ∏ni=1p(ai+1)∗ai2∗∏nj=1(aj+1)ai+1(mod109+6)i(mod109+7)
= ∏ni=1p(ai+1)∗ai(mod2∗(109+6))(mod2)∗∏nj=1(aj+1)ai+1(mod2∗(109+6))i(mod109+7)

注意10^9+6 无法求逆元，
所以先对2*(F-1)取模，再除2 处理
右边用 前缀积和后缀积处理

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
#include<vector>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=Pre[x];p;p=Next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=Next[p]) 
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define pb push_back
#define mp make_pair 
#define fi first
#define se second
#define MAXN (100000+10)
#define N (100000)
typedef long long ll;
typedef unsigned long long ull;
ll F = (1000000007LL);
ll mul(ll a,ll b,ll F){return a%F*b%F;}
ll add(ll a,ll b,ll F){return (a+b)%F;}
void upd(ll &a,ll b,ll F){a=(a%F+b%F)%F;}
int read()
{
    int x=0,f=1; char ch=getchar();
    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
    return x*f;
} 
ll pow2(ll a,ll b,ll F) {
    if (b==1) return a;
    if (!b) return 1;
    ll p=pow2(a,b/2,F);
    p=p*p%F;
    if (b&1) p=p*a%F;
    return p; 
}
//ll inv(ll a,ll F){return pow2(a,F-2);}
int n;
ll a[MAXN];
ll Ps[MAXN],Ss[MAXN]; 
vector<int> v[MAXN];
int main()
{
// freopen("d.in","r",stdin);
// freopen(".out","w",stdout);

    For(i,N) {
        int m=i;
        for(int d=2;d*d<=m;d++){
            while (m%d==0) {
                v[i].pb(d);
                m/=d;
            }
        }
        if (m>1) v[i].pb(m);        
    }
    while (cin>>n)
    {
        For(i,n) {
            a[i]=read();
        }
        a[1]=0;
        Fork(i,4,n){

            int m=v[i].size();
            if (m>1) {
                Rep(j,m) 
                    upd(a[v[i][j]],a[i],(F-1)*2);
                a[i]=0;
            }
        }
        a[1]=0;
        ll ans=1;
        Ps[0]=Ss[n+1]=1;
        For(i,n) Ps[i]=mul(Ps[i-1],(ll)a[i]+1,F-1);
        ForD(i,n) Ss[i]=mul(Ss[i+1],(ll)a[i]+1,F-1);

        Fork(i,2,n) {
            if (v[i].size()>1) continue; 
            ll b=1;
            if (a[i]&1) b=mul((a[i]+1)/2,a[i],F-1);
            else b=mul(a[i]+1,a[i]/2,F-1);

            b=mul(b,mul(Ps[i-1],Ss[i+1],F-1 ),F-1 );
            ans=mul(ans,pow2(i,b,F),F); 
        }
        cout<<ans<<endl;

    }
    return 0;
}