按照m之间的性质来决定采用什么方法

• 是否互质？

  • 互质——中国剩余定理

  • 不互质——拓展欧几里得 + 拓展中国剩余定理

• 互质情况下，是否均为质数？

  • 是—— 费马小定理求逆元

  • 否——拓展欧几里得求逆元

互质条件下的拓展欧几里得+中国剩余定理

题目(本题mi会爆long long，要么高精度、要么__int128，要么用拓展欧几里得)

代码

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef __int128 bt;

const int N = 11;
ll m[N], r[N];
int n;
ll ex_gcd(ll a, ll b, ll& x, ll& y)
{
    if(b == 0)
    {
        x = 1;
        y = 0;
        return a;
    }
    
    ll gcd = ex_gcd(b, a%b, x, y);
    ll tmp = x;
    x = y;
    y = tmp - a / b * y;
    
    return gcd;
}
ll crt(ll m[], ll r[])
{
    ll res = 0;
    
    bt mi = 1;
    for(int i = 1; i <= n; i++)
     mi *= m[i];
    
    ll x, y, M;
    for(int i = 1; i <= n; i++)
    {
        M = mi / m[i];
        ll gcd = ex_gcd(M, m[i], x, y);
        
        if(r[i] % gcd) return -1;
        
        x /= gcd;
        x = (x % (m[i] / gcd) + (m[i] / gcd)) % (m[i] / gcd);
        res = (res + r[i] * M % mi * x % mi) % mi;
    }
    
    return res;
}
int main()
{
    cin >> n;
    for(int i = 1; i <= n; i++)
        cin >> m[i] >> r[i];
    
    ll res = crt(m, r);
    cout << res;
}

无互质条件下的拓展欧几里得+拓展中国剩余定理

题目

代码

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 26;
ll m[N], r[N];
int n;
ll ex_gcd(ll a, ll b, ll&x, ll& y)
{
    if(b == 0)
    {
        x = 1;
        y = 0;
        return a;
    }
    ll gcd = ex_gcd(b, a % b, x, y);
    ll tmp = x;
    x = y;
    y = tmp - a / b * y;
    
    return gcd;
}
ll ex_crt(ll m[], ll r[])
{
    ll m1, m2, r1, r2, x, y;
    m1 = m[1], r1 = r[1];
    for(int i = 2; i <= n; i++)
    {
        m2 = m[i], r2 = r[i];
        ll gcd = ex_gcd(m1, m2, x, y);
        if((r2-r1) % gcd) return -1;
        x = x * (r2-r1) / gcd;
        x = (x % (m2 / gcd) + (m2 / gcd)) % (m2 / gcd);
        r1 = m1 * x + r1;
        m1 = m1 * m2 / gcd;
    }
    
    return (r1 % m1 + m1) % m1; 
}
int main()
{
    cin >> n;
    for(int i = 1; i <= n; i++)
        cin >> m[i] >> r[i];
        
    cout << ex_crt(m, r);
}