中国剩余定理

模板代码：

#include<bits/stdc++.h>
using namespace std;
using ll = long long;
#define fi first
#define se second
const ll mod=998244353;
const int N=2e5+10;
#define int ll


int n,m[300],r[300];
int exgcd(int a,int b,int &x,int &y){
	if(b==0){
		x=1,y=0;
		return a;
	}
	int d=exgcd(b,a%b,y,x);
	y-=a/b*x;
	return d;
}
ll crt(ll m[],ll r[]){
	ll M=1,ans=0;
	for(int i=1;i<=n;i++) M*=m[i];
	for(int i=1;i<=n;i++){
		ll c=M/m[i],x,y;
		exgcd(c,m[i],x,y);
		ans=(ans+r[i]*c*x%M)%M;
	}
	return (ans%M+M)%M;
}
void solve(){
	cin>>n;
	for(int i=1;i<=n;i++){
		cin>>m[i]>>r[i];
	}
	cout<<crt(m,r);
}

signed main(){
	ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    int t=1;
    //cin>>t;
	while (t--){
		solve();
	}
	return 0;
}