HDU 1573 X问题

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;

#include <iostream>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <string>

const int maxn = 500 + 20;
LL mod[maxn];
LL r[maxn];
LL gcd(LL n, LL m) {
if (n % m == 0) return m;
else return gcd(m, n % m);
}
LL lcm(LL n, LL m) {
return n / gcd(n, m) * m;
}
LL exgcd (LL a,LL mod,LL &x,LL &y) {
//求解a关于mod的逆元     ★：当且仅当a和mod互质才有用
if (mod==0) {
x=1;
y=0;
return a;//保留基本功能，返回最大公约数
}
LL g=exgcd(mod,a%mod,x,y);
LL t=x;    //这里根据mod==0  return回来后，
x=y;    //x,y是最新的值x2,y2,改变一下，这样赋值就是为了x1=y2
y=t-(a/mod)*y;    // y1=x2(变成了t)-[a/mod]y2;
return g;            //保留基本功能，返回最大公约数
}
bool get_min_number (LL a,LL b,LL c,LL &x,LL &y) {  //得到a*x+b*y=c的最小整数解
LL abGCD = gcd(a,b);
if (c % abGCD != 0) return false;//不能整除GCD的话，此方程无解
a /= abGCD;
b /= abGCD;
c /= abGCD;
LL tx,ty;
exgcd(a,b,tx,ty); //先得到a*x+b*y=1的解，注意这个时候gcd(a,b)=1
x = tx * c;
y = ty * c; //同时乘上c，c是约简了的。得到了一组a*x + b*y = c的解。
LL haveSignB = b;
if (b < 0) b = -b;   //避免mod负数啊，模负数没问题，模了负数后+负数就GG
x = (x % b + b) % b; //最小解
//    if (x == 0) x = b; //避免x = 0不是"正"整数  不要用这个，溢出
y = (c - a * x) / haveSignB;
return true;//true代表可以
}

void work () {
LL n, m;
cin >> n >> m;
for (int i = 1; i <= m; ++i) {
cin >> mod[i];
}
for (int i = 1; i <= m; ++i) {
cin >> r[i];
}
LL mm = mod[1];
LL ansx = 0;
LL rr = r[1];
for (int i = 2; i <= m; ++i) {
LL x, y;
bool ret = get_min_number(mm, -mod[i], r[i] - rr, x, y);
if (ret == false) {
printf("0\n");
return;
}
ansx = mm * x + rr;
mm = lcm(mm, mod[i]);
rr = ansx % mm;
}
int add = rr != 0;
if (rr > n) {
printf("0\n");
return ;
}
cout << (n - rr) / mm + add << endl;
return ;
}

int main() {
#ifdef local
freopen("data.txt","r",stdin);
#endif
int t;
cin >> t;
while (t--) work();
return 0;
}