[cf920G][容斥原理+二分]

https://codeforc.es/contest/920/problem/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 Copy

3
7 22 1
7 22 2
7 22 3      

output Copy

9
13
15      

input Copy

5
42 42 42
43 43 43
44 44 44
45 45 45
46 46 46      

output Copy

187
87
139
128
141
題意：求出與p互質并且>x的第k小數
題解：首先求出p所有的質因子，狀壓枚舉出質因子相乘的情況，比如000001表示隻有一個最小的質因子，則在1-y中有y/prime[1]個，再通過容斥即可求出與p不互質的個數，此時二分答案y即可      

1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define debug(x) cout<<#x<<" is "<<x<<endl;
 4 typedef long long ll;
 5 ll sol(ll x,vector<ll>&f){
 6     ll res=0;
 7     int n=(int)f.size();
 8     for(int i=1;i<(1<<n);i++)
 9     {
10         ll now=1,sgn=-1;
11         for(int j=0;j<n;j++)
12             if(i&(1<<j))now*=f[j],sgn*=-1;
13         res+=sgn*(x/now);
14     }
15     return x-res;
16 }
17 int main(){
18     int t;
19     scanf("%d",&t);
20     while(t--){
21         ll x,p,k;
22         scanf("%lld%lld%lld",&x,&p,&k);
23         vector<ll>g;
24         for(ll i=2;i*i<=p;i++){
25             if(p%i==0){
26                 g.push_back(i);
27                 while(p%i==0){
28                     p/=i;
29                 }
30             }
31         }
32         if(p>1)g.push_back(p);
33         ll r=1000000000000LL;
34         ll l=0;
35         k+=sol(x,g);
36         ll ans;
37         while(l<=r){
38             ll mid=(l+r)/2;
39             if(sol(mid,g)>=k){
40                 ans=mid;
41                 r=mid-1;
42             }
43             else{
44                 l=mid+1;
45             }
46         }
47         printf("%lld\n",ans);
48     }
49     return 0;
50 }      

View Code

轉載于:https://www.cnblogs.com/MekakuCityActor/p/10926041.html