D. Same GCDs

题目链接:D. Same GCDs

time limit per test:2 seconds
memory limit per test:256 megabytes
inputstandard input
outputstandard output

You are given two integers a and m. Calculate the number of integers x such that 0≤x<m and gcd(a,m)=gcd(a+x,m).
Note: gcd(a,b) is the greatest common divisor of a and b.
Input
The first line contains the single integer T (1≤T≤50) — the number of test cases.
Next T lines contain test cases — one per line. Each line contains two integers a and m (1≤a<m≤1010).
Output
Print T integers — one per test case. For each test case print the number of appropriate x-s.
input

3
4 9
5 10
42 9999999967

output

6
1
9999999966

Note
In the first test case appropriate x-s are [0,1,3,4,6,7].
In the second test case the only appropriate x is 0.

题目大意

给你两个数 a,m。从[a,a+m)中取出任意一个数x,使得gcd(x,m)=gcd(a,m)成立,问你这样的的x的个数一共有几个。

解题思路

我们设 p=gcd(a,m); 则p=gcd(x,m);然后同时除以p得1=gcd(x/p,m/p);这时候我们发现其实要求的的也就是[a/p,a/p+m/p)中与m/p互质的数有几个。同时减去a/p的,也就是[1,m/p)中于m/p互质的的个数,直接用欧拉函数就可以了;

代码

#include<bits/stdc++.h>
using namespace std;
#define ll long long

int main()
{
    int t;
    ll a,m;
    cin>>t;
    while(t--)
    {
        cin>>a>>m;
        ll p=__gcd(a,m);
        ll ans=m/p;
        ll q=ans;
        //欧拉函数
        for(ll i=2;i*i<=q;i++)
        {
            if(q%i==0) ans=ans-ans/i;
            while(q%i==0) q/=i;
        }
        if(q!=1) ans-=ans/q;
        cout<<ans<<"\n";
    }
    return 0;
}

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