Minimal Power of Prime HDU - 6623 2019 Multi-University Training Contest 4

Minimal Power of Prime
You are given a positive integer n > 1. Consider all the different prime divisors of n. Each of them is included in the expansion n into prime factors in some degree. Required to find among the indicators of these powers is minimal.
Input
The first line of the input file is given a positive integer T ≤ 50000, number of positive integers n in the file. In the next T line sets these numbers themselves. It is guaranteed that each of them does not exceed 10^18.
Output
For each positive integer n from an input file output in a separate line a minimum degree of occurrence of a prime in the decomposition of n into simple factors.
Sample Input

5
2
12
108
36
65536

Sample Output
1
1
2
2
16

题意：
求N以内素因子幂的最小值

分析：
一开始用Pollard_Rho交了3次，全TTTTTT了，tcl（太菜了），确实处理的方法有点暴力，如果那么简单，做的人就不会那么少了

看了别人家的孩子做的，原来这是有点东西啊，
先求1e4里面的素因子，这样求的是有1229个，为什么要求1e4个之内的呢，因为T组数据，题目中说 T ≤ 50000，那么这就是5e7的复杂度（需要暴力这些素数），可以在1s内飘过，然后还有一个地方就是，第1229个素数（9973^4）的四次方是9,892,436,613,211,441‬这个，第1230个素数是10007,也就是说第1229个素数之后最坏的情况是还有4个素因子相乘

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<string>
#include<cmath>
#include<cstring>
#include<set>
#include<queue>
#include<stack>
#include<map>
#define rep(i,a,b) for(int i=a;i<=b;i++)
typedef long long ll;
using namespace std;
const int N=1e5+10;
const ll INF=0x3f3f3f3f3f3f3f3f;
int maxn=1e5+10;
bool is_prime[20010];
ll prime[20010],tot;
void get_prime(){
    for(int i=2;i<=10000;i++){
        if(!is_prime[i]){
            prime[tot++]=i;
            for(int j=i+i;j<=10000;j+=i){
                is_prime[j]=1;
            }
        }
    }
}
int solve(ll n){
    int ans=(ll)1<<30;
    for(int i=0;i<tot;i++){//一个或者多个的
        if(n%prime[i]==0){
            int cnt=0;
            while(n%prime[i]==0){
                n/=prime[i];
                cnt++;
            }
            ans=min(ans,cnt);
            if(ans==1) return ans;//提前结束
        }
    }
    if(n!=1){
        ll k1= sqrt(n);
        if(k1*k1==n){//2个或者4个的
            ll k2=sqrt(k1);
            if(k2*k2==k1){
                return min(ans,4);
            }else
                return min(ans,2);
        }else{//三个或者1个的
            ll lb=0,ub=1000000;//1e6的三次方刚好是1e18，看n是不是某个数的三次方，用二分找会快一点
            while(ub>=lb){
                ll mid=(ub+lb)/2;
                if(mid*mid*mid==n)//找到了，说明剩下的数是一个可以开三次方的数
                    return min(ans,3);
                if(mid*mid*mid>n)
                    ub=mid-1;
                else
                    lb=mid+1;
              }
              return min(ans,1);//没有找到，那么就说明剩下的是一个大素数了
        }
    }else{
        return ans;
    }
}
int main()
{
    #ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
    #endif // ONLINE_JUDGE
    get_prime();
    int T;ll n;
    scanf("%d",&T);
    while(T--){
        scanf("%lld",&n);
        printf("%d\n",solve(n));
    }
    return 0;
}