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 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.
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).
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
直接java暴力
import java.math.*; import java.util.*; public class Main{ /** * @param args */ public static void main(String[] args) { Scanner cin=new Scanner(System.in); BigDecimal n; BigDecimal tmp=new BigDecimal("10"); BigDecimal tmpp=new BigDecimal("0"); MathContext mc = new MathContext(1000,RoundingMode.HALF_DOWN); int ans; while(cin.hasNext()){ ans=-1; n=cin.nextBigDecimal(); if(n.compareTo(tmpp)==0){ System.out.println("TAT");continue; } for(int j=1;j<=5;j++){ BigDecimal d = new BigDecimal(Math.sqrt(n.doubleValue()),mc); n=d; if(n.compareTo(tmp)<0){ if(n.intValue()==1){ ans=j;break; } } } if(ans!=-1)System.out.println(ans); else System.out.println("TAT"); } } }