uva 11752 The Super Powers

原题：

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 = 8 2 and

64 = 4 3 . You have to write a program that lists all super powers within 1 and 2 64 − 1 (inclusive).

Input

This program has no input.

Output

Print all the Super Power Numbers within 1 and 2 64 − 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

.

.

.

中文：

找出所有1到2^64-1的范围当中的一个数，这个数可以表示为a^x和b^y。没有输入

#include <bits/stdc++.h>
using namespace std;
typedef unsigned long long ull;
int p[20]={2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61};
int vis[70];
const ull maxn=(2<<64)-1;
set<ull> ans;

int main()
{
ios::sync_with_stdio(false);
ans.insert(1);
for(int i=0;i<18;i++)
vis[p[i]]=1;
for(ull i=2;;i++)
{
ull tmp=maxn;
int ind=-1;
while(tmp)
{
tmp/=i;
ind++;
}
if(ind<4)
break;
tmp=i;
for(ull j=2;j<=ind;j++)
{
tmp*=i;
if(!vis[j])
ans.insert(tmp);
}
}
for(auto x:ans)
cout<<x<<endl;

return 0;
}

解答：

只要找出一个数，让他的次幂数是一个合数即可。但是程序没写出来，看别人题解明白了。每次都用2^64-1不断去除一个数，除了多少次，就是这个数的次幂数。放入set中去重即可。