中国剩余定理

 

 

 

 

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
 LL exgcd(LL a, LL b, LL &x, LL &y)
{
    if (!b)
    {
        x = 1, y = 0;
        return a;
    }
    LL d = exgcd(b, a % b, y, x);
    y -= a / b * x;
    return d;
}
int main(){
    int n;
    cin>>n;
    LL a1,m1;
    cin>>a1>>m1;
    bool no=false;
    for (int i = 1; i < n; i ++ ){
        LL a2,m2;
        cin>>a2>>m2;
        LL k1,k2;
       LL d=exgcd(a1,a2,k1,k2);
        if((m2-m1)%d){
           no=true;
           break;
        }
            LL t=a2/d;
            k1=k1*(m2-m1)/d;
            k1=(k1%t+t)%t;
            m1=a1*k1+m1;
            a1=abs(a1/d*a2);
    }
      if (no) printf("-1\n");
    else printf("%lld\n", (m1 % a1 + a1) % a1);
    return 0;
}