[模板] 扩展中国剩余定理

将同余方程改写等式，用exgcd合并即可

#include<bits/stdc++.h>
#define int __int128
using namespace std;

int rd(){
	int ret=0,f=1;char c;
	while(c=getchar(),!isdigit(c))f=c=='-'?-1:1;
	while(isdigit(c))ret=ret*10+c-'0',c=getchar();
	return ret*f;
}

void exgcd(int A, int B, int &x, int &y) {
	if(!B) x = 1, y = 0;
	else exgcd(B, A%B, y, x), y -= (A/B) * x;
}

const int MAXN = 100005;

int n;

signed main(){
	n=rd();
	int preans=0,lcm=1;
	for(int i=1;i<=n;i++){
		int p,r,x,y;
		p=rd();r=rd();
		int g=__gcd(lcm,p);
		exgcd(lcm,p,x,y);
		if((r-preans)%g) return 0;
		int stp=p/g;
		x%=stp;
		if(x<0) x+=stp;
		x*=(r-preans)/g;
		preans+=x*lcm;
		lcm*=p/g;
		preans%=lcm;
		if(preans<0) preans+=lcm;
	}
	cout<<(long long)preans;
	return 0;
}