hdu 1573 X问题 ，模线性方程组

3
10 3
1 2 3
0 1 2
100 7
3 4 5 6 7 8 9
1 2 3 4 5 6 7
10000 10
1 2 3 4 5 6 7 8 9 10
0 1 2 3 4 5 6 7 8 9

1
0
3

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;
typedef long long ll;

ll ex_gcd(ll a, ll b, ll &x, ll &y){
	if(a==0 && b==0) return -1;
	if(b==0) {x=1; y=0; return a;}
	ll d = ex_gcd(b, a%b, y, x);
	y -= a/b*x;
	return d;
}

bool solve(int &m0, int &a0, int m, int a){
	ll y, x;
	ll g = ex_gcd(m0, m, x, y);
	if( abs(a-a0)%g ) return false;
	x *= (a-a0)/g;
	x %= m/g;
	a0 = (x*m0 + a0);
	m0 *= m/g;
	a0 %= m0;
	if(a0 < 0) a0 += m0;
	return true;
}

int mm[11], aa[11]; 
//模数为mm[i],余数为aa[i],X % mm[i] = aa[i]
// X = a0 + m0*t (0<=a0<m0)
bool MLES(int &m0, int &a0, int n){
	bool flag = true;
	m0 = 1;
	a0 = 0;
	for(int i=0; i<n; ++i)
		if( !solve(m0, a0, mm[i], aa[i]) ){
			flag = false;
			break;
		}
	return flag;
}

int main(){
	int T, n, m;
	freopen("in.txt","r",stdin);
	scanf("%d", &T);
	while(T--){
		scanf("%d%d", &n, &m);
		for(int i=0; i<m; ++i) scanf("%d", &mm[i]);
		for(int i=0; i<m; ++i) scanf("%d", &aa[i]);
		int m0, a0;
		// 0<(x=a0+m0*t)<=n
		if( !MLES(m0, a0, m) || n<a0)
			printf("0\n");
		else 
			printf("%d\n", (n-a0)/m0+1-(a0==0?1:0) );
	}
	return 0;
}