zoukankan      html  css  js  c++  java
  • HDU5878(打表)

    I Count Two Three

    Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 348    Accepted Submission(s): 184


    Problem Description

    I will show you the most popular board game in the Shanghai Ingress Resistance Team.
    It all started several months ago.
    We found out the home address of the enlightened agent Icount2three and decided to draw him out.
    Millions of missiles were detonated, but some of them failed.

    After the event, we analysed the laws of failed attacks.
    It's interesting that the i-th attacks failed if and only if i can be rewritten as the form of 2a3b5c7d which a,b,c,d are non-negative integers.

    At recent dinner parties, we call the integers with the form 2^a3^b5^c7^d "I Count Two Three Numbers".
    A related board game with a given positive integer n from one agent, asks all participants the smallest "I Count Two Three Number" no smaller than n.
     

    Input

    The first line of input contains an integer t (1t500000), the number of test cases. t test cases follow. Each test case provides one integer n (1n109).
     

    Output

    For each test case, output one line with only one integer corresponding to the shortest "I Count Two Three Number" no smaller than n.
     

    Sample Input

    10
    1
    11
    13
    123
    1234
    12345
    123456
    1234567
    12345678
    123456789
     

    Sample Output

    1
    12
    14
    125
    1250
    12348
    123480
    1234800
    12348000
    123480000
     

    Source

     
    枚举a,b,c,d,打表,二分查找
     1 //2016.9.17
     2 #include <iostream>
     3 #include <cstdio>
     4 #include <algorithm>
     5 #define N 10000
     6 #define ll long long 
     7 
     8 using namespace std;
     9 
    10 int arr[N];
    11 
    12 ll pow(ll a, ll b)//快速幂
    13 {
    14     ll ans = 1;
    15     while(b)
    16     {
    17         if(b & 1)ans *= a;
    18         a *= a;
    19         b>>=1;
    20     }
    21     return ans;
    22 }
    23 
    24 int main()
    25 {
    26     ll tmp; int cnt = 0;
    27     for(int a = 0; a < 31; a++)
    28     {
    29         for(int b = 0; b < 20; b++)
    30         {
    31             for(int c = 0; c < 14; c++)
    32             {
    33                 for(int d = 0; d < 12; d++)
    34                 {
    35                     tmp = pow(2, a)*pow(3, b);
    36                     if(tmp > 1e9)break;
    37                     tmp *= pow(5, c);
    38                     if(tmp > 1e9)break;
    39                     tmp *= pow(7, d);
    40                     if(tmp > 1e9)break;
    41                     arr[cnt++] = tmp;
    42                 }
    43             }
    44         }
    45     }
    46     sort(arr, arr+cnt);
    47     int T, n;
    48     scanf("%d", &T);
    49     while(T--)
    50     {
    51         scanf("%d", &n);
    52         int pos = lower_bound(arr, arr+cnt, n)-arr;
    53         printf("%d
    ", arr[pos]);
    54     }
    55 
    56     return 0;
    57 }
  • 相关阅读:
    Ubuntu 12.04下GAMIT10.40安装说明
    GAMIT 10.50在Ubuntu 12.04系统下的安装
    tomcat 5.5 动态加载类
    GAMIT 10.50在Ubuntu 12.04系统下的安装
    RHCE 系列(九):如何使用无客户端配置 Postfix
    Nginx+Keepalived(带Nginx监控脚本)
    黑马程序员_java08_多线程
    oracle 表类型变量的使用
    如何在win7系统中安装redis
    bzoj 2816: [ZJOI2012]网络(splay)
  • 原文地址:https://www.cnblogs.com/Penn000/p/5879942.html
Copyright © 2011-2022 走看看