Educational Codeforces Round 37 (Rated for Div. 2)G. List Of Integers 二分

题目链接：G. List Of Integers

题意：给你x,p,k，问你比x大的于p互质的第k个数.

题解：我们二分答案，如何判断呢，我们先的会算1~n于p互质的个数，不会的可以看这个，其他就很简单了。

#include<bits/stdc++.h>
#include<set>
#include<iostream>
#include<string>
#include<cstring>
#include<algorithm>
#define pb push_back
#define ll long long
#define PI 3.14159265
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#define eps 1e-7
typedef unsigned long long ull;
const int mod=1e9+7;
const int maxn=3e5+5;
const int root=1e6+7;
using namespace std;
int t;
ll x,pp,k,tmp;
vector<ll> p;
void init(int n)
{
    p.clear();
     for(ll i=2;i*i<=n;i++)
    {
        if(n%i==0)
        {
            p.pb(i);
            while(n%i==0)n/=i;
        }
    }if(n>1)p.pb(n);
}
ll slove(ll r)
{
    ll ans=0;
    for(ll msk=1;msk<(1<<p.size());msk++)
    {
        ll multi=1,bits=0;
        for(ll i=0;i<p.size();i++)
        {
            if(msk&(1<<i))++bits,multi*=p[i];
        }
        ll cnt=r/multi;
        if(bits&1)ans+=cnt;
        else ans-=cnt;
    }
    return r-ans;
}
bool judge(ll m)
{
    if(slove(m)-tmp>=k)
    {
        return true;
    }
    return false;
}
int main()
{
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lld %lld %lld",&x,&pp,&k);
        init(pp);
        ll l=x+1,r=1e12;
        ll ans;
        tmp=slove(x);
        while(l<=r)
        {
            ll m=(l+r)/2;
            if(judge(m))ans=m,r=m-1;
            else l=m+1;
        }
        printf("%lld
",ans);
    }
    return 0;
}