C++实现（同余方程组）中国剩余定理

#include
#include
using namespace std;
#define LL long long

/***********************************
中国剩余数定理
ai=x(mod mi)    求x
eg:1=x(mod 11) 2=x(mod 7)
最后求得为满足条件的最小值
************************************/
LL ex_gcd(LL a,LL b,LL &x,LL &y)
{
    if(b==0)
    {
        x=1;y=0;
        return a;
    }
    LL r=ex_gcd(b,a%b,x,y);
    LL t=x;
    x=y;
    y=t-a/b*y;
    return r;
}

///n个mi互质
//const LL maxn = 20;
//LL a[maxn], m[maxn], n;
//LL CRT(LL a[], LL m[], LL n)
//{
//    LL M = 1;
//    for (int i = 0; i < n; i++) M *= m[i];
//    LL ret = 0;
//    for (int i = 0; i < n; i++)
//    {
//        LL x, y;
//        LL tm = M / m[i];
//        ex_gcd(tm, m[i], x, y);
//        ret = (ret + tm * x * a[i]) % M;
//    }
//    return (ret + M) % M;
//}

///n个mi不互质
const LL maxn = 1000;
LL a[maxn], m[maxn], n;
LL CRT(LL a[], LL m[], LL n) {
    if (n == 1) {
        if (m[0] > a[0]) return a[0];
        else return -1;
    }
    LL x, y, d;
    for (int i = 1; i < n; i++) {
        if (m[i] <= a[i]) return -1;
        d = ex_gcd(m[0], m[i], x, y);
        if ((a[i] - a[0]) % d != 0) return -1;  //不能整除
        LL t = m[i] / d;
        x = ((a[i] - a[0]) / d * x % t + t) % t;
        a[0] = x * m[0] + a[0];
        m[0] = m[0] * m[i] / d;
        a[0] = (a[0] % m[0] + m[0]) % m[0];
    }
    return a[0];
}

int main ()
{
   LL T;
   while( ~scanf("%lld",&T) )
   {
       for(int i=0; i 
  
转自：https://blog.csdn.net/u010468553/article/details/38346195