现在方程是x=b（mod c）

pi.first = c, pi.second = b;

代码按照第一篇题解思路模拟而来/

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#define int long long

using namespace std;

inline int read(){
	register int x = 0, f = 1; register char ch = 0;
	while(!isdigit(ch)){
		if(ch == '-') f = -1;
		ch = getchar();
	}
	while(isdigit(ch)){
		x = (x << 1) + (x << 3) + (ch ^ '0');
		ch = getchar();
	}
	return x * f;
}

int n;
vector<pair<int, int> > vec;

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

void exgcd(int a, int b, int &x, int &y){
	if(!b) {
		x = 1; y = 0; return;
	}
	exgcd(b, a % b, y, x); y -= (a / b) * x;
}

signed main(){
	n = read();
	for(int i = 1; i <= n; ++ i){
		pair<int,int> pi;
		pi.first = read(); pi.second = read();
		vec.push_back(pi);
	}
	while(vec.size() != 1){
		pair<int, int>pi1, pi2;
		pi1 = vec.back(); vec.pop_back();
		pi2 = vec.back(); vec.pop_back();
		int h = gcd(pi1.first, pi2.first);
//		cout << pi1.first << ' ' << pi1.second << endl << pi2.first << ' ' << pi2.second << endl;
		int p1 = pi1.first / h, p2 = pi2.first / h;
		int r = (pi2.second - pi1.second) / h;
		int lcm = p1 * p2 * h;
		int k1, k2, x;
		exgcd(p1, p2, k1, k2);
		x = ((((k1 % p2 + p2) % p2 * r % lcm) % lcm * pi1.first % lcm + pi1.second) % lcm + lcm) % lcm;
//		cout << x << endl;
		pi1.first = lcm;
		pi1.second = x;
		vec.push_back(pi1);
	}
	pair<int ,int> pi = vec.back(); int x, y;
	exgcd(1, pi.first, x, y);
	
	cout << pi.second;
}