We all know the Super Powers of this world and how they manage to get advantages in political warfare or even in other sectors.  But this is not a political platform and so we  will talk about a different kind     of super powers — “The Super Power Numbers”.  A positive number is said to be super power when it   is the power of at least two different positive integers. For example 64 is a super power as 64 = 82 and  64 = 43. You have to write a program that lists all super powers within 1 and 264 − 1 (inclusive).

 

Input

This program has no   input.

 

Output

Print all the Super Power Numbers within 1 and 264 − 1.  Each line contains a single super power  number and the numbers are printed in ascending  order.

Note:  Remember that there are no input for this problem.  The sample output is only a partial solution.

 

Sample Input

 

 

Sample Output

1

16

64

81

256

512

.

.

.

只要对于某个底数指数为合数即可
主要是找出对于每个指数,底数所能到达的上限
ull maxn=-1;
一是对于每个指数i 判断a^i能整除a^(i-1)且a^i/a^(i-1) =a，找的满足的最大值（如下面本人代码）;
二是用对数判断 对于每个指数i找到iloga<64log2的最大的a
三是枚举底数 用最大值maxn除底数,找到底数所能到达的最大指数
#include <iostream>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <cstdio>
#include <vector>
#include <map>
#include <cstdlib>
using namespace std;
typedef unsigned long long ull;
vector<ull> ans;
int limit[100007];
bool prime[100007];
const ull maxn=-1;
int main()
{
//    freopen("D:\\Administrator\\Desktop\\1.txt","w",stdout);
    for(int i=4;i<64;i++)
    {
        for(ull j=1;j<=maxn;j++)
        {
            ull x=1,mm;
            int f=0;
            for(int k=0;k<i-1;k++)
            {
                x*=j;
                mm=x*j;
                if(x==0||mm%x!=0||mm/x!=j)
                {
                    f=1;
                    break;
                }
            }
            if(f)
            {
                limit[i]=j;
                break;
            }
        }
    }
    for(int i=2;i<=64;i++)
        if(!prime[i])
        {
            for(int j=2*i;j<=64;j+=i)
                prime[j]=true;
        }
    for(int i=4;i<64;i++)
    {
        if(prime[i])
        {
            for(int j=2;j<limit[i];j++)
            {
                ull x=1;
                for(int k=0;k<i;k++)
                    x*=j;
                ans.push_back(x);
            }
        }
    }
    ans.push_back(1);
    sort(ans.begin(),ans.end());
    ans.erase(unique(ans.begin(),ans.end()),ans.end());
    for(int i=0;i<ans.size();i++)
        cout<<ans[i]<<endl;
    return 0;
}