伟大的中国剩余定理

实现代码

long long ex_gcd(long long a, long long b,long long& x,long long &y)
{
	long long m;
	if (b == 0)
	{
		x = 1;
		y = 0;
		return a;
	}
	else
	{
		m = ex_gcd(b, a%b, x, y);
		long long t = x;
		x = y;
		y = t - a / b*y;
	}
	return m;
}

int flag = 0;
long long a1, a2, n1, n2, k1, k2;

void china_rem(long long a[], long long n[],int num)
{
	flag = 0;
	a1 = a[0];
	n1 = n[0];
	for (int i = 1; i < num; ++i)
	{
		a2 = a[i];
		n2 = n[i];
		long long d = ex_gcd(n1, n2, k1, k2);
		long long c = a2 - a1;
		if (c%d != 0)
		{
			flag = 1;
			return;
		}
		else
		{
			long long t = abs(n2 / d);
			k1 = (((c / d)*k1)%t + t) % t;//最小正整数解,这样子最后求得的a1才是最小正数
			a1 += k1*n1;
			n1 = n1 / d*n2;
			a1 = (a1%n1 + n1) % n1;//尤其要注意，再次取余，不然可能会出错
		}
	}
}