【模板】扩展中国剩余定理（EXCRT）

Code:

#include <cstdio>
#include <algorithm>
#define N 100005
#define ll long long
#define ull unsigned long long
#define setIO(s) freopen(s".in","r",stdin)
using namespace std;
int n;
ll arr[N],brr[N];
ll mult(ll x,ll y,ll mod)
{
	ll tmp=(long double)x/mod*y;
	return ((ull)x*y-(ull)tmp*mod+mod)%mod;
}
ll qpow(ll base,ll k,ll mod)
{
	ll tmp=1;
	for(;k;k>>=1,base=mult(base,base,mod)) if(k&1) tmp=mult(tmp,base,mod);
	return tmp;
}
ll exgcd(ll a,ll b,ll &x,ll &y)
{
	if(!b)
	{
		x=1,y=0;
		return a;
	}
	ll gcd=exgcd(b,a%b,x,y),tmp=x;
	x=y,y=tmp-a/b*y;
	return gcd;
}
ll Excrt()
{
	int i,j;
	ll ans=brr[1],M=arr[1];
	for(i=2;i<=n;++i)
	{
		ll a=M,b=arr[i],x,y,c=brr[i]-ans,gcd;
		gcd=exgcd(a,b,x,y),b=abs(b/gcd);
		x=mult(x,c/gcd,b);
		ans+=M*x;
		M*=arr[i]/gcd;
		ans=(ans%M+M)%M;
	}
	return ans;
}
int main()
{
	int i,j;
	// setIO("input");
	scanf("%d",&n);
	for(i=1;i<=n;++i) scanf("%lld%lld",&arr[i],&brr[i]);
	printf("%lld\n",Excrt());
	return 0;
}