codeforces Round680 C. Division 题解

题目

Oleg's favorite subjects are History and Math, and his favorite branch of mathematics is division.

To improve his division skills, Oleg came up with \(t\) pairs of integers \(p_i\) and \(q_i\) and for each pair decided to find the greatest integer \(x_i\), such that:

  • \(p_i\) is divisible by \(x_i\);
  • \(x_i\) is not divisible by \(q_i\).

Oleg is really good at division and managed to find all the answers quickly, how about you?

Input

The first line contains an integer \(t\) \((1≤t≤50)\) — the number of pairs.

Each of the following \(t\) lines contains two integers \(p_i\) and \(q_i\) \((1≤ p_i≤10^{18}; 2≤q_i≤10^9)\)— the \(i-th\) pair of integers.

Output

Print \(t\) integers: the \(i-th\) integer is the largest \(x_i\) such that \(p_i\) is divisible by \(x_i\), but \(x_i\) is not divisible by \(q_i\).

One can show that there is always at least one value of \(x_i\)i satisfying the divisibility conditions for the given constraints.

Example

input

3
10 4
12 6
179 822

output

10
4
179

Note

For the first pair, where \(p_1=10\) and \(q_1=4\), the answer is \(x_1=10\), since it is the greatest divisor of \(10\) and \(10\) is not divisible by \(4\).

For the second pair, where \(p_2=12\) and \(q_2=6\), note that

  • \(12\) is not a valid \(x_2\), since \(12\) is divisible by \(q_2=6\);
  • \(6\) is not valid \(x_2\) as well: \(6\) is also divisible by \(q_2=6\).

The next available divisor of \(p_2=12\) is \(4\), which is the answer, since \(4\) is not divisible by \(6\).

题意

找一个最大的\(x\)，满足\(p\ \%\ x == 0 \ and\ x\ \%\ q != 0\)。

思路

质因数分解

  • \(p\ \%\ q\ !=\ 0\)， \(x = p\)
  • \(p\ \%\ q\ =\ 0\) , 那么\(p\)一定包含\(q\)的所有质因数分解的结果。

举例：

\(p = 12\ \ q = 6\)
\(p = 2^2 * 3^1\) \(q = 2^1 +3^1\)

要使\(p\ \%\ q\ !=\ 0\), 只要使 \(p\) 将质因数\(2\)降幂到\(0\)(也就是q的质因数\(2\)的次幂之下)，或者将\(3\) 的幂降到\(0\)。

所以，我们只需要枚举\(q\)的质因子, 找权值最小的，即为答案。

代码

#include <bits/stdc++.h>
using namespace std;
#define ll unsigned long long
#define IOS ios_base::sync_with_stdio(0); cin.tie(0);cout.tie(0);
const int maxn = 2e5 + 10;
int ans[maxn];

void solve(){
    ll p, q;
    cin >> p >> q;
    if(p % q) { //p不能被q整除，答案就是p
        cout << p << endl;
        return;
    }
    ll ans = 0;
    for (ll i = 1; i * i <= q; i++){
        if(q % i) continue; // 不是q的因子
        //i 和 q/i 都是因子
        ll t = p;
        if(i != 1){
            while(t % q == 0) t /= i; 
            ans = max(ans, t);
        }
        t = p;
        while(t % q == 0) t /= (q / i);
        ans = max(ans, t);
    }
    cout << ans << endl;
}
int main(){
    IOS; int t; cin >> t;
    while(t--){
        solve();
    }
    return 0;
}