zoukankan      html  css  js  c++  java
  • HDU 5752 Sqrt Bo

    Sqrt Bo

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
    Total Submission(s): 699    Accepted Submission(s): 325


    Problem Description
    Let's define the function f(n)=⌊√n.

    Bo wanted to know the minimum number y which satisfies fy(n)=1.

    note:f1(n)=f(n),fy(n)=f(fy-1(n))

    It is a pity that Bo can only use 1 unit of time to calculate this function each time.

    And Bo is impatient, he cannot stand waiting for longer than 5 units of time.

    So Bo wants to know if he can solve this problem in 5 units of time.
     
    Input
    This problem has multi test cases(no more than 120).

    Each test case contains a non-negative integer n(n<10100).
     
    Output
    For each test case print a integer - the answer y or a string "TAT" - Bo can't solve this problem.
     
    Sample Input
    233
    233333333333333333333333333333333333333333333333333333333
     
    Sample Output
    3
    TAT
     
    Source
     

    解析:题意为给定一个数n(0 ≤ 10100),不断地开方并向下取整,问经过多少次操作后n等于1。当操作次数大于5时输出"TAT",当操作次数小于或等于5时输出操作的次数。因为有5次的限制,我们可以找到一个分界点。易知这个分界点为232:当n≥232时,输出"TAT"(当n==0时也输出"TAT"),其他情况与2的对应次幂比较即可。

    #include <bits/stdc++.h>
    #define ll long long
    
    char s[105];
    
    int main()
    {
        while(~scanf("%s", s)){
            int len = strlen(s);
            if(len>10){
                puts("TAT");
                continue;
            }
            ll sum = 0;
            for(int i = 0; i < len; ++i)
                sum = sum*10+s[i]-'0';
            if(sum >= 1ll<<32 || sum == 0) puts("TAT");
            else if(sum >= 1ll<<16) puts("5");
            else if(sum >= 1ll<<8) puts("4");
            else if(sum >= 1ll<<4) puts("3");
            else if(sum >= 1ll<<2) puts("2");
            else if(sum >= 1ll<<1) puts("1");
            else puts("0");
        }
        return 0;
    }
    
  • 相关阅读:
    解决一起web 页面被劫持的案例
    django rest framwork教程之外键关系和超链接
    django restframwork 教程之authentication权限
    Puppet nginx+passenger模式配置
    Django restframwork教程之类视图(class-based views)
    使用emplace操作
    C++中的显示类型转换
    整数加法
    在不知道学生人数和每个学生课程数量的情况下对学生的平均成绩排序
    树的高度
  • 原文地址:https://www.cnblogs.com/inmoonlight/p/5710761.html
Copyright © 2011-2022 走看看