P4777 【模板】扩展中国剩余定理（EXCRT）

#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;

const int N=100010;

int n;
long long aa[N],bb[N];

long long ksc(long long a,long long b,long long mod) {
	long long res=0;
	while(b>0) {
		if(b&1)
			res=(res+a)%mod;
		a=(a+a)%mod;
		b>>=1;
	}
	return res;
}

long long exgcd(long long a,long long b,long long &x,long long &y) {
	if(b==0) {
		x=1;
		y=0;
		return a;
	}
	long long gcd=exgcd(b,a%b,x,y);
	long long tp=x;
	x=y;
	y=tp-a/b*y;
	return gcd;
}

long long work() {
	long long x,y,k;
	long long M=bb[1],ans=aa[1];
	for(int i=2; i<=n; i++) {
		long long a=M,b=bb[i],c=((aa[i]-ans)%b+b)%b;
		long long gcd=exgcd(a,b,x,y),bg=b/gcd;
		if(c%gcd!=0)
			return -1;
		x=ksc(x,c/gcd,bg);
		ans+=x*M;
		M*=bg;
		ans=(ans%M+M)%M;
	}
	return (ans%M+M)%M;
}

int main () {
	freopen("excrt.in","r",stdin);
	freopen("excrt.out","w",stdout);
	scanf("%d",&n);
	for(int i=1; i<=n; i++) {
		scanf("%lld",&bb[i]);
		scanf("%lld",&aa[i]);
	}
	printf("%lld\n",work());
	return 0;
}