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 2^64 -1. Each line contains a single super power number and the numbers are printed in ascending order.
Sample Output
1
16
64
81
256
512
.
.
.
题意：求超级幂。一个数n能表示成ai^bi，a,b至少存在两组时，n为超级幂。输出2^64-1内的所有超级幂。
解析：底数在[2,1<<16]内，指数是64以内的合数。底数为a的幂不会溢出的上限为ceil(64/((log(a)/log(2)))-1。
一开始以为底数应该是素数，导致wa了很多次。
代码如下：
# include<iostream>
# include<cstdio>
# include<cstring>
# include<set>
# include<vector>
# include<cmath>
# include<algorithm>
using namespace std;
const int N=105;
int vis[N];
unsigned long long mypow(int a,int b)
{
if(b==0)
return 1;
if(b==1)
return a;
unsigned long long res=mypow(a,b>>1);
res*=res;
if(b&1)
res*=a;
return res;
}
void get_prim()
{
int i,j;
fill(vis,vis+N,0);
for(i=2;i<N;++i){
if(vis[i])
continue;
for(j=i+i;j<N;j+=i)
vis[j]=1;
}
}
int main()
{
int i,j;
get_prim();
set<unsigned long long>s;
s.insert(1);
set<unsigned long long>::iterator it;
for(i=2;i<(1<<16);++i){
for(j=4;j<=ceil(64/(log(i)/log(2)))-1;++j){
if(vis[j]){
s.insert(mypow(i,j));
}
}
}
for(it=s.begin();it!=s.end();++it)
printf("%llu
",*it);
return 0;
}