扩展中国剩余定理

#include <bits/stdc++.h>
using namespace std;
#define ll long long
const int N = 1e5 + 5;

ll n;
ll ai[N], bi[N];

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

ll kmul(ll a, ll b, ll p)
{
    ll ans = 0;
    while(b)
    {
        if(b & 1) ans = (ans + a) % p;
        a = (a + a) % p;
        b >>= 1;
    }
    return ans;
}

ll excrt()
{
    ll x, y, k;
    ll M = bi[1], ans = ai[1]; //第一个方程的解特判
    for(int i = 2; i <= n; i++)
    {
        ll a = M, b = bi[i], c = (ai[i] - ans % b + b) % b; //ax≡c(mod b)
        ll gcd = exgcd(a, b, x, y), bg = b / gcd;
        if(c % gcd) return -1; //判断是否无解
        x = kmul(x, c / gcd, bg);
        ans += x * M; //更新前k个方程组的答案
        M *= bg; //M为前k个m的lcm
        ans = (ans % M + M) % M;
    }
    return (ans % M + M) % M;
}