求解~

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 thisproblem.

 
Sample Input
233
233333333333333333333333333333333333333333333333333333333
 

Sample Output
3
TAT

AC代码：

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<queue>
#include<algorithm>
#include<time.h>
using namespace std;

const int maxn = 2005;
const int N=100007;

#define lson rt<<1///rt*2
#define rson rt<<1|1///rt*2+1

char s[N];

int main()
{
    int len,i;

    while(scanf("%s", s) != EOF)
    {
        long long d=0;
        len=strlen(s);
        if(len>10)
        {
            printf("TAT
");
            continue ;
        }
        for(i=0; i<len; i++)
            d=d*10+s[i]-'0';


        if(d>=65536&&d<4294967296)
            printf("5
");
        else if(d>=256&&d<65536)
            printf("4
");
        else if(d>=16&&d<256)
            printf("3
");
        else if(d>=4&&d<16)
            printf("2
");
        else if(d>1&&d<4)
            printf("1
");
        else if(d==1)
            printf("0
");
        else
            printf("TAT
");
    }
    return 0;
}

Wa代码：

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<queue>
#include<algorithm>
#include<time.h>
using namespace std;

const int maxn = 2005;
const int N=1000007;

#define lson rt<<1///rt*2
#define rson rt<<1|1///rt*2+1

char s[N];

int main()
{
    int len,i;

    while(scanf("%s", s) != EOF)
    {
        long long d=0;
        len=strlen(s);
        if(len>10)
        {
            printf("TAT
");
            continue ;
        }
        for(i=0; i<len; i++)
            d=d*10+s[i]-'0';

        if(d<4294967296)
        {
            if(d>=65536)
                printf("5
");
            else if(d>=256)
                printf("4
");
            else if(d>=16)
                printf("3
");
            else if(d>=4)
                printf("2
");
            else if(d>1)
                printf("1
");
            else if(d==1)
                printf("0
");
        }
        else
            printf("TAT
");
    }
    return 0;
}

很搞不懂，谁来解释一下 if else 1号 和 if else2号 的区别，~~~~(>_<)~~~~