Sqrt Bo Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 980    Accepted Submission(s): 452 Problem Description Let's define the function f(n)=⌊n√⌋. Bo wanted to know the minimum number y which…

Sqrt Bo 题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5752 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…

Sqrt Bo Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 2221    Accepted Submission(s): 882 Problem Description Let's define the function f(n)=⌊n−−√⌋. Bo wanted to know the minimum number y wh…

Sqrt Bo 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5752 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…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5752 题目大意: 定义f(n)=⌊√n⌋,fy(n)=f(fy-1(n)),求y使得fy(n)=1.如果y>5输出TAT.(n<10100) 题目思路: [模拟] 5层迭代是232,所以特判一下层数是5的,其余开根号做.注意数据有0. 队友写的. #include<stdio.h> #include<string.h> #include<math.h> int…

题目:传送门. 题意:一个很大的数n,最多开5次根号,问开几次根号可以得到1,如果5次还不能得到1就输出TAT. 题解:打表题,x1=1,x2=(x1+1)*(x1+1)-1,以此类推.x5是不超过long long的,判断输出即可. #include <iostream> #include <cstdio> #include <cstring> #include <cmath> using namespace std; typedef long long…

#include<bits/stdc++.h> using namespace std; ], comp; ]; int main(){ arr[] = 1LL; ; i < ; i ++) arr[i] = (arr[i - ] + ) * (arr[i - ] + ) - ; while(~scanf("%s", str)){ int len = strlen(str); ){ printf("TAT\n"); }else{ sscanf(st…

Sqrt Bo Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 2222    Accepted Submission(s): 883 Problem Description Let's define the function f(n)=⌊√n⌋ . Bo wanted to know the minimum number y whi…

Sqrt Bo Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 145    Accepted Submission(s): 72 Problem Description Let's define the function f(n)=⌊n−−√⌋. Bo wanted to know the minimum number y whic…

5/11 2016 Multi-University Training Contest 3官方题解 2016年多校训练第三场 老年选手历险记 暴力 A Sqrt Bo(CYD) 题意:问进行多少次开根号向下取整能使给定的值为1,若为5次以上,则输出TAT: 思路: 有5次这个限制,得知值最大为10的11次方左右,存下后运算即可,注意初始值为0时运算的次数是大于5次的,输出TAT: 代码: #include <bits/stdc++.h> using namespace std; char st…