https://codeforces.com/contest/1445/problem/C
To improve his division skills, Oleg came up with tt pairs of integers pipi and qiqi and for each pair decided to find the greatest integer xi, such that:
- pi is divisible by xi;
- xi is not divisible by qi.
pi and qi (1≤pi≤1e18; 2≤qi≤1e9)
思路:从质因数分解的角度出发,那么可以发现,如果两个数不能整除,直接就是pi.
如果能整除,那么他们的质因子种类都是同的。只不过质因数的次幂不同。那么我们只要让pi其一个质因子次幂降到qi的对应质因子次幂的少一个就能满足不整除。至于选哪一个枚举即可。
(由于时间复杂度把不断整除算成了sqrt(n)导致卡了好久-.-)
#include<iostream>
#include<vector>
#include<queue>
#include<cstring>
#include<cmath>
#include<map>
#include<set>
#include<cstdio>
#include<algorithm>
#define debug(a) cout<<#a<<"="<<a<<endl;
using namespace std;
const int maxn=1e5;
typedef long long LL;
LL qpow(LL a,LL n)
{
LL ans=1;
while(n)
{
if(n&1) ans*=a;
a*=a;
n>>=1;
}
return ans;
}
int main(void)
{
cin.tie(0);std::ios::sync_with_stdio(false);
LL t;cin>>t;
while(t--){
LL p,q;cin>>p>>q;
if(p%q!=0){
cout<<p<<endl;
}
else{
LL x=p;
LL ans=0x3f3f3f3f3f3f3f3f;
for(LL i=2;i*i<=q;i++)
{
if(q%i) continue;
LL c1=0,c2=0;
while(q%i==0)
{
c1++;
q/=i;
}
while(p%i==0)
{
c2++;
p/=i;
}
ans=min(ans,qpow(i,c2-c1+1));
}
if(q>1)
{
LL i=q;
LL c1=0,c2=0;
while(q%i==0)
{
c1++;
q/=i;
}
while(p%i==0)
{
c2++;
p/=i;
}
ans=min(ans,qpow(i,c2-c1+1));
}
cout<<x/ans<<endl;
}
}
return 0;
}