2020牛客寒假算法基础集训营6 E.立方数（唯一分解定理 素数筛）

https://ac.nowcoder.com/acm/contest/3007/E

放下题解

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 32640;
ll vis[maxn*100],prime[maxn],prime_3[maxn];
ll tot = 0;
void getPrime(){
    memset(vis,0,sizeof(vis));
    for(ll i=2;i<=maxn;i++){
        if(!vis[i])prime[tot++]=i,prime_3[tot-1] = i*i*i;
        for(ll j=0;j<tot;j++){
            if(prime[j]*i>maxn)break;
            vis[prime[j]*i]=1;
            if(i%prime[j]==0)break;
        }
    }
}

int main(){
    getPrime();
    int t;scanf("%d",&t);
    while(t--){
        ll n;
        ll ans = 1;
        scanf("%lld",&n);
        for(int i = 0;i<tot&&prime[i]<=n;i++){
            while(n%prime_3[i] == 0){
                ans *=prime[i];
                n/=prime_3[i];
            }
            while(n%prime[i] == 0){
                n/=prime[i];
            }
        }
        int L = 1, R = 1000000;
        while(L<=R){//二分试除
            int mid = (L+R)/2;
            if((ll)mid*mid*mid<n) L = mid + 1;
            else R = mid - 1;
        }
        if((ll)L*L*L == n) ans*=L;
        printf("%lld
",ans);
    }
    return 0;
}