hdu 5752 Sqrt Bo

keepcoral

于 2018-03-31 11:10:07 发布

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分类专栏：思维题

本文链接：https://blog.csdn.net/keepcoral/article/details/79767083

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Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 5 Accepted Submission(s): 5

Problem Description

Let's define the function .

Bo wanted to know the minimum number which satisfies .

note:

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 ).

Each test case contains a non-negative integer .

Output

For each test case print a integer - the answer or a string "TAT" - Bo can't solve this problem.

Sample Input

   233233333333333333333333333333333333333333333333333333333333 
 

Sample Output

   3  TAT 
 

一开始我也没发现规律...还以为要用高精度平方来做，但实际上是思维题，因为我们只需要找到开五次方等于1最大的数就可以了，sqrt2=1,sqrt3=1,sqrt4=2,sqrt15=2,sqrt16=3,所以sqrt(2^32)=6，而sqrt(2^32-1)=5,这恰好就是它的最大的数，因此，只需要判断给出的n是否大于等于sqrt(2^32)就可以了，如果大于等于，那么直接输出TAT，如果不是，那么还需要判断n！=0,因为sqrt(0)!=1,然后直接对n操作就可以了

#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cmath>
#include<cstring>
using namespace std;
int a[5009];
long long num=4294967296-1;
int main()
{
    char s[109];
    long long n,i;
    while(~scanf("%s",s))
    {
        n=strlen(s);
        if(n>10||s[0]=='0')
        {
            printf("TAT\n");
        }
        else
        {
            long long sum=0;
            int count=0;
            for(i=0;i<=n-1;i++)
            {
                sum=sum*10+s[i]-'0';
            }
            if(sum>num)
            {
                printf("TAT\n");
                continue;
            }
            while(sum!=1)
            {
                sum=(long long )sqrt(sum);
                count++;
            }
            printf("%d\n",count);
        }
    }
}