Codeforces Round #499 (Div. 2) E. Border(贝祖定理)

题解思路:

对于每个啊a[i]先对k取余;

贝祖定理可知若 a1*x1+a2*x2+.....+an*xn==c;

则 c|gcd(a1,a2,a3.....an);

所以ans=(gcd*(0->k-1))%k;

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<map>
#include<queue>
#include<stack>
#include<set>

#define ll long long
#define int long long
#define mem(a,b) memset(a,b,sizeof(a))

const int maxn=1e5+50;

using namespace std;

int gcd(int a,int b)
{
    return b==0?a:gcd(b,a%b);
}

set<int>q;

#undef int
int main(){
#define int long long
    int n,k,g;
    cin>>n>>k;
    cin>>g;
    for(int i=2;i<=n;i++)
    {
        int now;
        cin>>now;
        g=gcd(g,now);
    }
    for(int i=0;i<k;i++)
    {
        q.insert((g*i)%k);
    }
    cout<<q.size()<<endl;
    set<int>::iterator iter;
    for(iter=q.begin();iter!=q.end();)
    {
        cout<<*iter;
        ++iter!=q.end()?cout<<" ":cout<<endl;
    }
    return 0;
}