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{egin{array}{l} {x equiv a_{1}left(mod m_{1} ight)} \ {x equiv a_{2}left(mod m_{2} ight)} \ {cdots} \ {x equiv a_{n}left(mod m_{n} ight)} end{array} ight. ]

有整数解。并且在模(M = m_{1}*m_{2}..m_{n})下的解是唯一的，解为

[oldsymbol{x} equivleft(oldsymbol{a}_{1} M_{1} M_{1}^{-1}+a_{2} M_{2} M_{2}^{-1}+ldots+a_{k} M_{k} M_{k}^{-1} ight) mod 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 + M) % M;
}
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)

[left{egin{array}{l} {x equiv a_{1}left(mod m_{1} ight)} \ {x equiv a_{2}left(mod m_{2} ight)} \ {cdots} \ {x equiv a_{n}left(mod m_{n} ight)} end{array} ight. ]

对于(egin{cases}x equiv a_1(mod m_1)\xequiv a_2(mod m_2) end{cases})

进行变化有

(egin{cases} x=a_1+k_1 imes m_1 \ x=a_2 +k_2 imes m_2 end{cases})

(a_1 + m_1 imes k_1 = a_2 + m_2 imes k_2\m_1k_1 = (a_2 - a_1) + m_2k_2)

方程有解当且仅当(gcd(m_1,m_2)|(a_2 - a_1))

设(gcd(m_1,m_2) = d)

(frac{m_1k_1}{d} = frac{a_2 - a_1}{d} + frac{m_2k_2}{d} \ frac{m_1}{d}k_1 equiv frac{a_2 - a_1}{d}(mod frac{m_2}{d}))

两边同时除(frac{m_1}{d})

(k_1 equiv inv(frac{m_1}{d},frac{m_2}{d}) * frac{c_2 - c_1}{d} (mod frac{m_2}{d}) \ k_1 = inv(frac{m_1}{d},frac{m_2}{d}) * frac{a_2 - a_1}{d} + frac{m_2}{d} * y)

把原式(x = a_1 + k_1m_1)带入

(x = inv(frac{m_1}{d},frac{m_2}{d}) * frac{a_2 - a_1}{d} * m_1 + yfrac{m_1m_2}{d} + a_1 \ x equiv inv(frac{m_1}{d},frac{m_2}{d}) * frac{a_2 - a_1}{d} * m_1 + a_1(mod frac{m_1m_2}{d}))

那么答案就是(x equiv a(mod m))

其中(a = (inv (frac{m_1}{d}, frac{m_2}{d}) * frac{a_2 - a_1}{d}) \%frac{m_2}{d} *m_1 + a_1\m = frac{m_1m_2}{d})

不断地进行两两合并，然后利用线性同余方程求出这两个同余方程的组的x，然后继续合并

#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++){
        ex_gcd(M,m[i],d,x,y);
        if((ans - a[i]) % d)return -1;
        ll c = ((a[i] - ans) % m[i] + m[i]) % m[i];
        ll mod = m[i]/d;
        x = mull(x,c/d,mod);
        ans += x * M;
        M *= mod;
        ans = (ans % M + M) % M;
    }
    return ans;
}
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;
}