中国剩余定理公式例题：从经典孙子问题到现代算法应用

def extended_gcd(a, b):
    if a == 0:
        return b, 0, 1
    else:
        g, y, x = extended_gcd(b % a, a)
        return g, x - (b // a)  y, y
def modinv(a, m):
    g, x, y = extended_gcd(a, m)
    if g != 1:
        raise Exception('Modular inverse does not exist')
    else:
        return (x % m + m) % m
def chinese_remainder_theorem(n, a):
    """
    n: 模数列表 [m1, m2, ..., mk]
    a: 余数列表 [a1, a2, ..., ak]
    返回: 最小非负解 x
    """
    if len(n) != len(a):
        raise Exception('n and a must have same length')
    M = 1
    for m in n:
        M = m
    x = 0
    for i in range(len(n)):
        M_i = M // n[i]
        y_i = modinv(M_i, n[i])
        x += a[i]  M_i  y_i
    return x % M
测试用例：孙子算经原题
n = [3, 5, 7]
a = [2, 3, 2]
print("最小解:", chinese_remainder_theorem(n, a))  # 输出: 23
        

#include <iostream>
#include <vector>
using namespace std;
typedef long long ll;
// 扩展欧几里得算法
ll extended_gcd(ll a, ll b, ll &x, ll &y) {
    if (b == 0) {
        x = 1;
        y = 0;
        return a;
    }
    ll x1, y1;
    ll d = extended_gcd(b, a % b, x1, y1);
    x = y1;
    y = x1 - y1  (a / b);
    return d;
}
// 模逆元
ll modinv(ll a, ll m) {
    ll x, y;
    ll g = extended_gcd(a, m, x, y);
    if (g != 1) return -1; // 无逆元
    return (x % m + m) % m;
}
// 中国剩余定理
ll crt(const vector<ll>& n, const vector<ll>& a) {
    ll M = 1;
    for (ll m : n) M = m;
    ll x = 0;
    for (int i = 0; i < n.size(); i++) {
        ll M_i = M / n[i];
        ll y_i = modinv(M_i, n[i]);
        x += a[i]  M_i  y_i;
    }
    return (x % M + M) % M;
}
int main() {
    vector<ll> n = {3, 5, 7};
    vector<ll> a = {2, 3, 2};
    cout << "最小解: " << crt(n, a) << endl;
    return 0;
}