Educational Codeforces Round 37 (Rated for Div. 2) G

G. List Of Integers

time limit per test

5 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Let's denote as L(x, p) an infinite sequence of integers y such that gcd(p, y) = 1 and y > x (where gcd is the greatest common divisor of two integer numbers), sorted in ascending order. The elements of L(x, p) are 1-indexed; for example, 9, 13 and 15 are the first, the second and the third elements of L(7, 22), respectively.

You have to process t queries. Each query is denoted by three integers x, p and k, and the answer to this query is k-th element of L(x, p).

Input

The first line contains one integer t (1 ≤ t ≤ 30000) — the number of queries to process.

Then t lines follow. i-th line contains three integers x, p and k for i-th query (1 ≤ x, p, k ≤ 106).

Output

Print t integers, where i-th integer is the answer to i-th query.

Examples

input

37 22 17 22 27 22 3

output

input

542 42 4243 43 4344 44 4445 45 4546 46 46

output

18787139128141

题意 q个询问  大于x，第k个与p互质的数解析 对于一个数 mid 我们可以容斥算出1-mid 与 p互质的数有多少，所以二分答案就可以了。AC代码

#include <bits/stdc++.h>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define all(a) (a).begin(), (a).end()
#define fillchar(a, x) memset(a, x, sizeof(a))
#define huan printf("\n")
#define debug(a,b) cout<<a<<" "<<b<<" "<<endl
#define ffread(a) fastIO::read(a)
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int maxn=1e4+10;
const ll mod=1000000007;
ll yinzi[maxn],cnt;
void euler(ll n)
{cnt=0;ll a=n;for(ll i=2; i*i<=a; i++){if(a%i==0){yinzi[cnt++]=i;while(a%i==0)a/=i;}}if(a>1)yinzi[cnt++]=a;
}
ll solve(ll n)
{ll ans=0;for(ll i=0; i<(1<<cnt); i++){ll temp=1,jishu=0;for(ll j=0; j<cnt; j++){if(i&(1<<j))temp=temp*yinzi[j],jishu++;}if(jishu==0)continue;if(jishu&1)ans+=n/temp;elseans-=n/temp;}return ans;
}
int main()
{ll t,n,m,k;scanf("%lld",&t);while(t--){scanf("%lld%lld%lld",&m,&n,&k);euler(n);ll ans1=m-solve(m);ll l=m+1,r=1e7;while(l<=r){ll mid=(l+r)/2;ll cur=mid-solve(mid)-ans1;if(cur<k)l=mid+1;elser=mid-1;}printf("%lld\n",r+1);}
}

转载于:https://www.cnblogs.com/stranger-/p/9811509.html

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