Sqrt Bo

Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 0????Accepted Submission(s): 0




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
  
  
  
  

  
  

?

题目意思：

对一个数开方，能在五次之内取整得到0的就输出次数；不能或者是0的话就输出TAT。。

解题思路：

这题longlong暴力就能过，队友很快就过了~然而我当时没想到，只想到了大数……QwQ

吭哧吭哧改了半天……才过…╮(╯▽╰)╭

longlong的代码很简单我就不写了，下面给出我写的大数的…这个模板说不定以后用的上呢~2333

#include <iostream>
using namespace std;
#include <stdlib.h>
#include <string.h>
#define MAXN 1000000
int j,l,size,num,x[MAXN];
char ch[MAXN],ans[MAXN];
void sqrt(char *str)//大数求平方
{
    double i,r,n;
    size=strlen(str);
    if (size%2==1)
    {
        n=str[0]-48;
        l=-1;
    }
    else
    {
        n=(str[0]-48)*10+str[1]-48;
        l=0;
    }
    r=0,num=0;
    while (true)
    {
        i=0;
        while (i*(i+20*r)<=n)
            ++i;
        --i;
        n-=i*(i+20*r);
        r=r*10+i;
        x[num]=(int)i;
        ++num;
        l+=2;
        if (l>=size)
            break;
        n=n*100+(double)(str[l]-48)*10+(double)(str[l+1]-48);
    }


}
int main()
{
    while (cin>>ch)
    {
        int flag=0;
        sqrt(ch);
        if((x[num-1]==1)&&num==1)//注意判断条件
        {
            cout<<"1"<<endl;
            flag=1;
            goto A;
        }
        for(int i=2; i<=5; ++i)//最多计算五次
        {
            for (j=0; j<num; ++j)
                ans[j]='0'+x[j];//int转char
            ans[num]='\0';//注意末尾别忘了加结束符
            //cout<<ans<<endl;
            sqrt(ans);//开方
          if((x[num-1]==1)&&num==1)
            {
                cout<<i<<endl;//输出次数
                flag=1;//标记
                break;
            }
        }
       A: if(!flag)
            cout<<"TAT"<<endl;//超过五次或者为0时
    }
    return 0;
}