• The Super Powers


    Time Limit: 1000MS   Memory Limit: Unknown   64bit IO Format: %lld & %llu

    []   [Go Back]   [Status]  

    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
    ",*it);
    67 }
    View Code
  • 相关阅读:
    Android 基于Message的进程间通信 Messenger完全解析
    Android获取TextView显示的字符串宽度
    关于移动App的五个提问
    Android 结合实例学会AsyncTask的用法
    提高 Android 代码质量的4个工具
    高效开发Android App的10个建议
    移动5年 Android生态系统的演进
    最受欢迎的游戏引擎集结号:跨平台篇
    Java程序员转Android开发必读经验分享
    8 个最优秀的 Android Studio 插件
  • 原文地址:https://www.cnblogs.com/ERKE/p/3700280.html
Copyright © 2020-2023  润新知