The Super Powers

The Super Powers

Time Limit: 1000MS

Memory Limit: Unknown

64bit IO Format: %lld & %llu

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Description

 

A

The Super Powers

 

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 64 81 256 512 . . .

 

 

 1 #include<stdio.h>
 2 #include<string.h>
 3 #include<set>
 4 #include<math.h>
 5 #include<iostream>
 6 #include<algorithm>
 7 using namespace std;
 8 #define  ll unsigned long long
 9 int a[100]= {0},nu;
10 void init()
11 {
12     int i,j;
13     nu=0;
14     for(i=2; i<100; i++)
15     {
16         if(!a[i])
17         {
18             j=i*i;
19             while(j<100)
20             {
21                 a[j]=1;
22                 j+=i;
23             }
24         }
25     }
26     j=0;
27     for(i=4; i<65; i++)if(a[i])a[j++]=i;
28     //for(i=0; i<j; i++)cout<<a[i]<<endl;
29     nu=j;
30 }
31 ll powLL(ll x, ll y)
32 {
33     ll ans=1;
34     while(y)
35     {
36         if(y&1)
37         ans*=x;
38         x*=x;
39         y>>=1;
40     }
41     return ans;
42 }
43 int main()
44 {
45     init();
46     int i,j,size;
47     ll  maxa=~0ULL,n;
48     //cout<<maxa<<endl;
49     set<ll>aa;
50     aa.clear();
51     aa.insert(1);
52     for(i=2;;i++)
53     {
54         size=0;
55         n=maxa;
56         while(n>=i) n/=i,size++;
57         if(size<4) break;
58         for(j=0;j<nu;j++)
59         {
60             if(a[j]<=size)
61             aa.insert(powLL(i, a[j]));
62             else break;
63         }
64     }
65     for(set<ll>::iterator it=aa.begin();it!=aa.end();it++)
66     printf("%llu\n",*it);
67 }

View Code

 

转载于:https://www.cnblogs.com/ERKE/p/3700280.html