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  • 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
");
        continue;
    }

    LL n = 0;
    sscanf(str,"%I64d", &n);

    LL cmps = 1LL<<32;
    if(n >= cmps || !n) {
        printf("TAT
");
        continue;
    }

    int cnt = 0;
    while(n != 1){
        n = sqrt(n);
        cnt++;
    }

    printf("%d
", cnt);
}

return 0;

    }
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  • 原文地址：https://www.cnblogs.com/Sunshine-tcf/p/5709459.html
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