11752 - The Super Powers
Time limit: 1.000 seconds
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 Input | Partial Judge Output |
No input for this problem | 1 16 81 512 |
用set实现，也可以用数组和sort&unique实现（更快）
完整代码：
/*0.036s*/
#include<cstdio>
#include<cmath>
#include<set>
#define LL unsigned long long
using namespace std;
const int index[100] =
{
4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25, 26,
27, 28, 30, 32, 33, 34, 35, 36, 38, 39, 40, 42, 44, 45, 46,
48, 49, 50, 51, 52, 54, 55, 56, 57, 58, 60, 62, 63, 64 /// 多算一个
};///合数表
set<LL> ans;
int main(void)
{
LL i, maxi = 1 << 16; /// 底数的上限
int j, maxindex; /// 指数的上限
LL cur;
ans.insert(1);
for (i = 2; i < maxi; i++)
{
cur = i * i; /// i用LL类型的原因
cur *= cur;
ans.insert(cur);
maxindex = ceil(64 * log(2) / log(i)) - 1; /// 指数的上限
for (j = 1; index[j] <= maxindex; ++j)
{
cur *= (index[j] - index[j - 1] == 1 ? i : i * i);
ans.insert(cur);
}
}
for (set<LL>::iterator iter = ans.begin(); iter != ans.end(); ++iter)
printf("%llu\n", *iter);
return 0;
}
