中国剩余定理

中国剩余定理

对于一组同余方程,\(m_{1},m_{2}...m_{n}两两互质\)
\[ \left\{\begin{array}{l}{x \equiv a_{1}\left(\bmod m_{1}\right)} \\{x \equiv a_{2}\left(\bmod m_{2}\right)} \\{\cdots} \\{x \equiv a_{n}\left(\bmod m_{n}\right)}\end{array}\right. \]
有整数解。并且在模\(M = m_{1}*m_{2}..m_{n}\)下的解是唯一的，解为
\[ \boldsymbol{x} \equiv\left(\boldsymbol{a}_{1} M_{1} M_{1}^{-1}+a_{2} M_{2} M_{2}^{-1}+\ldots+a_{k} M_{k} M_{k}^{-1}\right) \bmod M \]
其中\(M_{i} = \frac{M}{m_{i}},M^{-1}_{i}\)为\(M_{i}\)模\(m_{i}\)的逆元

#include <iostream>
#include <cstdio>
#define ll long long
using namespace std;
const int maxn = 1e5 + 5;
void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){
    if(b == 0){
        d = a,x = 1,y = 0;
        return;
    }
    ex_gcd(b,a%b,d,y,x);
    y -= x*(a/b);
}
ll china(ll a[],ll m[],int n){//x = a (mod m)
    ll M = 1,y,d,x = 0;
    for(int i = 0; i < n; i++)M *= m[i];
    for(int i = 0; i < n; i++){
        ll w = M/m[i];
        ex_gcd(m[i],w,d,d,y);
        x = (x + a[i] * w * y) % M;
    }
    if((x + M) % M == 0) return M;
    return x > 0? x:x + M;//x = 是M倍数时特判
}
int main(){
    int n;
    cin >> n;
    ll a[maxn],m[maxn];
    for(int i = 0; i < n; i++){
        cin >> a[i] >> m[i];
    }
    cout << china(a,m,n) << endl;
    return 0;
}

扩展中国剩余定理

当\(m_{1},m_{2}...m_{n}不一定两两互质时\)

\(x = a_{1} + m_{1}x_{1}\\x = a_{2}+m_{2}x_{2}\)

那么有\(m_{1}x_{1}+m_{2}x_{2} = a_{2}-a_{1}\)

只需要求出一个最小的x使得\(x = a_{1}+m_{1}x_{1}=a_{2}+m_{2}x_{2}\)

即\(x_{1}和x_{2}尽可能小，利用扩展欧几里得求出最小x_{1}，x_{2}然后带回去得到x的特解x'\)

\(x的通解一定是x'+lcm(m_{1}+m_{2})*k\)

那么
\[ \mathrm{x} \equiv x^{\prime}(\bmod l \operatorname{cm}(m 1, m 2)) \]

#include <iostream>
#include <cstdio>
#define ll long long
using namespace std;
const int maxn = 1e5 + 5;
void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){
    if(b == 0){
        d = a,x = 1,y = 0;
        return;
    }
    ex_gcd(b,a%b,d,y,x);
    y -= x * (a/b);
}
ll mull(ll a,ll b,ll mod){
    ll ans = 0;
    while(b){
        if(b & 1)ans = (ans + a) % mod;
        a = (a + a) % mod;
        b >>= 1;
    }
    return ans;
}
ll ex_china(ll a[],ll m[],int n){//x = a(mod m)
    ll M = m[0];
    ll ans = a[0];
    ll d,x,y;
    for(int i = 1; i < n; i++){
        ll c = ((a[i] - ans) % m[i] + m[i]) % m[i];
        ex_gcd(M,m[i],d,x,y);
        //if(c % d)return -1;//不存在解的情况
        ll mod = m[i]/d;
        x = mull(x,c/d,mod);
        ans += x * M;
        M *= mod;
        ans = (ans % M + M) % M;
    }
    return ans > 0?ans:ans + M;//注意ans是M倍数时输出M
}
int main(){
    int n;
    ll a[maxn],m[maxn];
    cin >> n;
    for(int i = 0; i < n; i++){
        cin >> a[i] >> m[i];
    }
    cout << ex_china(a,m,n) << endl;
    return 0;
}

来源：https://www.cnblogs.com/Emcikem/p/12234167.html