[Educational Codeforces Round 81 (Rated for Div. 2)] D. Same GCDs (数论，因子分解，容斥定理)

D. Same GCDs

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given two integers aa and mm. Calculate the number of integers xx such that 0≤x<m0≤x<m and gcd(a,m)=gcd(a+x,m)gcd(a,m)=gcd(a+x,m).

Note: gcd(a,b)gcd(a,b) is the greatest common divisor of aa and bb.

Input

The first line contains the single integer TT (1≤T≤501≤T≤50) — the number of test cases.

Next TT lines contain test cases — one per line. Each line contains two integers aa and mm (1≤a<m≤10101≤a<m≤1010).

Output

Print TT integers — one per test case. For each test case print the number of appropriate xx-s.

Example

input

Copy

3
4 9
5 10
42 9999999967

output

Copy

6
1
9999999966

Note

In the first test case appropriate xx-s are [0,1,3,4,6,7][0,1,3,4,6,7].

In the second test case the only appropriate xx is 00.

题意：

T组数据，

每一组数据给定两个整数a和m

问有多少个x满足：

\(0 \le x < m\) 和\(\gcd(a, m) = \gcd(a + x, m)\)

思路：

设\(b=\gcd(a,m)\),那么问题等同于问：

区间\([a,a+m-1]\) 中有多少个数x使\(\gcd(x,m)=b\)

又因为\(\gcd(x/b,m/b)=1\) , 所以问题可以转化为 :

区间\([a/b,(a+m-1)/b]\) 中有多少个数x使\(\gcd(x,m/b)=1\)

即区间\([a/b,(a+m-1)/b]\)中有多少个数x与m/b互质。

我们知道这是一个因子分解+容斥定理的经典问题。

代码：

#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 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) { if (a == 0ll) {return 0ll;} a %= MOD; 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 long long readll() {long long tmp = 0, fh = 1; char c = getchar(); while (c < '0' || c > '9') {if (c == '-') fh = -1; c = getchar();} while (c >= '0' && c <= '9') tmp = tmp * 10 + c - 48, c = getchar(); return tmp * fh;}
inline int readint() {int tmp = 0, fh = 1; char c = getchar(); while (c < '0' || c > '9') {if (c == '-') fh = -1; c = getchar();} while (c >= '0' && c <= '9') tmp = tmp * 10 + c - 48, c = getchar(); return tmp * fh;}
const int maxn = 1000010;
const int inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
vector<ll> list;
void  breakdown(ll x)
{
    list.clear();
    for (ll i = 2; i * i <= x; ++i) {
        int cnt = 0;
        while (x % i == 0) {
            ++cnt;
            x /= i;
        }
        if (cnt) {
            list.push_back(i);
        }
    }
    if (x > 1) {
        list.push_back(x);
    }
}

int main()
{
    //freopen("D:\\code\\text\\input.txt","r",stdin);
    //freopen("D:\\code\\text\\output.txt","w",stdout);
    int t;
    t = readint();
    ll a, m;
    while (t--)
    {
        a = readll();
        m = readll();
        ll b = gcd(a, m);
        breakdown(m / b);
        ll l = a;
        ll r = a + m - 1;
        l /= b;
        r /= b;
        ll ans = 0;
        int len = list.size();
        for (int i = 0; i < (1 << len); ++i) {
            int cnt = 0;
            ll pp = 1ll;
            for (int j = 0; j < len; ++j) {
                if (i & (1 << j)) {
                    ++cnt;
                    pp *= list[j];
                }
            }
            ans += (r / pp - (l - 1) / pp) * ((cnt & 1) ? -1 : 1);
        }
        printf("%lld\n", ans );
    }
    return 0;
}