codeforces - 326B - Duff in Love（练习）

B. Duff in Love

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Duff is in love with lovely numbers! A positive integer x is called lovely if and only if there is no such positive integer a > 1 such that a2 is a divisor of x.

Malek has a number store! In his store, he has only divisors of positive integer n (and he has all of them). As a birthday present, Malek wants to give her a lovely number from his store. He wants this number to be as big as possible.

Malek always had issues in math, so he asked for your help. Please tell him what is the biggest lovely number in his store.

Input

The first and only line of input contains one integer, n (1 ≤ n ≤ 1012).

Output

Print the answer in one line.

Sample test(s)

input

10

output

10

input

12

output

6

Note

In first sample case, there are numbers 1, 2, 5 and 10 in the shop. 10 isn't divisible by any perfect square, so 10 is lovely.

In second sample case, there are numbers 1, 2, 3, 4, 6 and 12 in the shop. 12 is divisible by 4 = 22, so 12 is not lovely, while 6 is indeedlovely.

这题RE的我够心塞，RE了两次！！wa我就认了吧，竟然会RE  QAQ心情是沉痛的。

题意：就是给一个n，输出他的最大的约数，而且这个约数不能被n其他约数的平方整除。

如果是质数的话，那就是n本身。如果是合数，扫一遍n的约数会超时，所以扫了n的开方次约数，然后n%约数== 0， 则也是n的约数，这样复杂度就降下来啦。只是判断的时候光顾着看平方后的范围了，没注意扫到n也是要停的QAQ

#include <iostream>#include <algorithm>#include <cstdio>#include <cmath>#include <cstring>#include <cstdlib>#include <iostream>#include <stack>#include <queue>#define INF 0x3f3f3f3fusing namespace std;long long a[10000005];int main(){  long long n;  while(scanf("%I64d", &n) != EOF)  {    long long k = 0;    long long ans;    for(long long i = 1; i * i <= n; i++)    {      if(n % i == 0)        a[k++] = i;    }    if(k == 1)      ans = n;    else    {      long long kk = k;      for(int i = 0; i < kk; i++)      {        if((n % a[i] == 0) && ((n / a[i]) > a[kk - 1]))          a[k++] = n / a[i];      }      sort(a, a + k);      for(long long i = k - 1; i >= 0; i--)      {        long long flag = 0;        for(long long j = 1; a[j] * a[j] <= a[i] && j < k; j++)   //最开始没有写j<k  QAQ  RE的好难过，还蠢蠢的找了半天错误T T        {          long long m = a[j] * a[j];          if(a[i] % m == 0)          {            flag = 1;            break;          }        }        if(flag == 0)  //flag = 0 就是这个约数不是任何约数平方的倍数啦，就是答案喵        {          ans = a[i];          break;        }      }    }    printf("%I64d\n", ans);  }  return 0;}