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

37 22 17 22 27 22 3

output

Copy

91315

input

Copy

542 42 4243 43 4344 44 4445 45 4546 46 46

output

Copy

18787139128141

题目链接：

https://codeforces.com/contest/920/problem/G

题意：

有t组询问，对于每一组询问，

给你三个整数x，p，k

问有在大于x的整数中，与p互质的第k小的数y是哪个？

思路：

对于每一组询问我们在区间\([x+1,1e9]\) 这个区间内，二分答案y

同时容斥定律可以求得区间\([l,r]\) 中与一个数num互质的数个数。——知识点[1]

那么我们可以求区间\([x+1,y]\)中与p互质的数个数与k比较，然后进行转移区间即可。

先筛出\(1e6\) 内的所有质数，然后\(log(p)\) 的时间复杂度去唯一分解询问中的p，然后二进制枚举+容斥定律辅助二分即可。

不会的话，建议先学一下知识点1，再来解决本题。

code:

#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <cmath>

#include <queue>

#include <stack>

#include <map>

#include <set>

#include <vector>

#include <iomanip>

#define ALL(x) (x).begin(), (x).end()

#define sz(a) int(a.size())

#define rep(i,x,n) for(int i=x;i<n;i++)

#define repd(i,x,n) for(int i=x;i<=n;i++)

#define pii pair<int,int>

#define pll pair<long long ,long long>

#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)

#define MS0(X) memset((X), 0, sizeof((X)))

#define MSC0(X) memset((X), '\0', sizeof((X)))

#define pb push_back

#define mp make_pair

#define fi first

#define se second

#define eps 1e-6

#define gg(x) getInt(&x)

#define chu(x) cout<<"["<<#x<<" "<<(x)<<"]"<<endl

#define du3(a,b,c) scanf("%d %d %d",&(a),&(b),&(c))

#define du2(a,b) scanf("%d %d",&(a),&(b))

#define du1(a) scanf("%d",&(a));

using namespace std;

typedef long long ll;

ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}

ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}

ll powmod(ll a, ll b, ll MOD) {a %= MOD; if (a == 0ll) {return 0ll;} ll ans = 1; while (b) {if (b & 1) {ans = ans * a % MOD;} a = a * a % MOD; b >>= 1;} return ans;}

void Pv(const vector<int> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%d", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("\n");}}}

void Pvl(const vector<ll> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%lld", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("\n");}}}

inline void getInt(int *p);

const int maxn = 1000010;

const int inf = 0x3f3f3f3f;

/*** TEMPLATE CODE * * STARTS HERE ***/

// const int maxn = 1e7 + 50;

bool noprime[maxn + 50];

vector <int> p;

int getPrime()

{

    // 华丽的初始化

    memset(noprime, false, sizeof(noprime));

    p.clear();

    int m = (int)sqrt(maxn + 0.5);

    // 优化的埃筛

    for (int i = 2; i <= m; i++) {

        if (!noprime[i]) {

            for (int j = i * i; j <= maxn; j += i) {

                noprime[j] = true;

            }

        }

    }

    // 把素数加到vector里

    for (int i = 2; i <= maxn; i++) {

        if (!noprime[i]) {

            p.push_back(i);

        }

    }

    //返回vector的大小

    return p.size();

}

std::vector<ll> v;

void breakdown(ll n, ll len)

{

    int pos = 0;

    for (int i = 0; 1ll * p[i]*p[i] <= n && i < len; i++) {

        if ( n % p[i] == 0) {

            v.push_back(p[i]);

            while (n % p[i] == 0) {

                n /= p[i];

            }

        }

    }

    if ( n > 1) {

        v.push_back(n);

    }

}

int x, pw, k;

int len ;

ll solve(ll l, ll r)

{

    int maxstate = (1 << len) - 1;

    ll ans = 0ll;

    l--;

    for (int i = 0; i <= maxstate; ++i) {

        int num = 0;

        ll p = 1ll;

        for (int j = 0; j < len; ++j) {

            if (i & (1 << j)) {

                num++;

                p *= v[j];

            }

        }

        ans += (r / p - l / p) * ((num & 1) ? -1ll : 1ll);

    }

    return ans;

}

int main()

{

    //freopen("D:\\code\\text\\input.txt","r",stdin);

    //freopen("D:\\code\\text\\output.txt","w",stdout);

    int n;

    int w = getPrime();

    du1(n);

    while (n--) {

        du3(x, pw, k);

        v.clear();

        breakdown(pw, w);

        len = sz(v);

        ll l = x + 1ll;

        ll r = 1e8;

        ll mid;

        ll ans;

        while (l <= r) {

            mid = (l + r) >> 1;

            if (solve(x + 1ll, mid) >= k) {

                ans = mid;

                r = mid - 1;

            } else {

                l = mid + 1;

            }

        }

        printf("%lld\n", ans );

    }

    return 0;

}

inline void getInt(int *p)

{

    char ch;

    do {

        ch = getchar();

    } while (ch == ' ' || ch == '\n');

    if (ch == '-') {

        *p = -(getchar() - '0');

        while ((ch = getchar()) >= '0' && ch <= '9') {

            *p = *p * 10 - ch + '0';

        }

    } else {

        *p = ch - '0';

        while ((ch = getchar()) >= '0' && ch <= '9') {

            *p = *p * 10 + ch - '0';

        }

    }

}