HDU 5752 Sqrt Bo （数论）

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

3
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;
}
posted @ 2016-07-26 23:52  Sunshine_tcf  阅读(332)  评论(0)    收藏  举报