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HDU 5752 Sqrt Bo （数论）

工程师路杨技术 2022年11月16日

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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 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

TAT

Source

2016 Multi-University Training Contest 3

##题意：

给出一个数(1e100)，问能否在开5次平方以内收敛到1.
若能则输出次数，否则输出TAT.

##题解：

首先找一个边界：恰好开5次平方不能到1的.很容易找出是2^32.
由于输入数据很大，所以要用字符串读入，长度大于10的串直接输出TAT.
长度合适的串用sscanf读出数据，先与2^32比较，再进行开平方操作.
要注意0的情况：0输出TAT.

##代码：
“` cpp
#include
#include
#include
#include
#include
#include
#include
#include
#include
#define LL long long
#define eps 1e-8
#define maxn 700
#define mod 100000007
#define inf 0x3f3f3f3f
#define IN freopen(“in.txt”,”r”,stdin);
using namespace std;

char str[120];

int main(int argc, char const *argv[])

{

//IN;

while(scanf("%s", str) != EOF)
{
    int len = strlen(str);
    if(len > 10) {
        printf("TAT\n");
        continue;
    }    LL n = 0;
    sscanf(str,"%I64d", &n);    LL cmps = 1LL<<32;
    if(n >= cmps || !n) {
        printf("TAT\n");
        continue;
    }    int cnt = 0;
    while(n != 1){
        n = sqrt(n);
        cnt++;
    }    printf("%d\n", cnt);
}return 0;