2022 Nioke winter vacation algorithm basic training camp 4, subsequence weight product (enumeration, Euler power reduction)

Portal

idea:

We can observe that after sorting , if the maximum and minimum values ​​are determined, the number in the middle has no effect on the result. Therefore, in this way, the maximum and minimum values ​​can be enumerated, and the middle part is reduced in At the same time, turn zero into an integer , then enumerate the length of the interval, combine the bases with the same power, and divide the two numbers by the inverse each time.

Euler's power reduction:

Request $a^b\%p$process, if$b$ is a relatively large number (may be in the form of a power), you can let$b$ pair$(p-1)$ Taking the modulo does not affect the correctness of the final answer (provided that$a$ and$p$ is relatively prime), this modulo is called Euler's power reduction (according to Fermat's little theorem, when$a$ and$When\; p$ is relatively prime, there is$a^{p-1}\%p=1$）

Code:

#include<bits/stdc++.h>
using namespace std;
#define ll long long
ll a[200010];
int mod=1e9+7;

ll power(ll a,ll b,ll mod){
    
    
    a%=mod;
    ll res=1;
    while(b){
    
    
        if(b&1)res=res*a%mod;
        b>>=1;
        a=a*a%mod;
    }
    return res;
}

ll inv(ll x,ll mod){
    
    
    return power(x,mod-2,mod);
}

int main(){
    
    
    int n;
    cin>>n;
    ll res=1,cnt=1,tmp=1;
    for(int i=0;i<n;i++){
    
    
        cin>>a[i];
        tmp=tmp*a[i]%mod*a[i]%mod;
    }
    sort(a,a+n);
    ll l=0,r=n-1;
    //子序列长度为1的情况
    res=tmp;
    while(l<n-1){
    
    
        tmp=tmp*inv(a[l++],mod)%mod*inv(a[r--],mod)%mod;
        res=res*power(tmp%mod,power(2,l-1,mod-1),mod)%mod%mod;
    }
    cout<<res;
}