屠龙勇士题解（系数不为一的同余方程组解法）
1.前言
还是对扩展中国剩余定理不熟，稍微变了一下就蒙了……
2.题解
首先用各种方法找到每条龙对应的剑。
根据题意，可以列出一下方程
\[atk_ix \equiv a_i \pmod {p_i}
\]
发现我们平时的扩展中国剩余定理有一个美妙的要求 \(atk_i = 1\)，所以我们要拼尽全力去满足这个要求。
我们先不考虑必须把龙的生命值打成非正数这个条件。
我们要表达原方程中 \(x\) 的取值，\(x\) 可取的条件是 \(\exists y \in Z， 满足 atk_ix + p_i y = a_i\)，则我们用扩展欧几里得解出 \(x\) 的一组特解\((x')\)后，可得 \(x \in \{x''\mid x'' = x' + k \frac{p_i}{gcd (p_i, atk_i)},k \in Z \}\)，所以 \(x \equiv x' \pmod {\frac{p_i}{gcd (p_i, atk_i)}}\)，现在，他不就是一个系数为一的同余方程了吗？
然后，我们记用扩展中国剩余定理合并后的同余方程组变为了 \(x \equiv b \pmod {m}\)，则 \(x \in \{x' \mid x' = b + km, k \in Z \}\)，则我们只需要在这个解集中找到一个满足要求(将所有龙的生命值打成非整数)的数，即在解集中找到最小的大于等于\(\max {\lceil \frac{a_i}{atk_i} \rceil}\)的数即可。
#include <set>
#include <cmath>
#include <cstdio>
#include <iostream>
using namespace std;
#define LL long long
#define ULL unsigned long long
#define PLL pair <LL, LL>
#define MP(x,y) make_pair (x, y)
template <typename T> void read (T &x) { x = 0; T f = 1;char tem = getchar ();while (tem < '0' || tem > '9') {if (tem == '-') f = -1;tem = getchar ();}while (tem >= '0' && tem <= '9') {x = (x << 1) + (x << 3) + tem - '0';tem = getchar ();}x *= f; return; }
template <typename T> void write (T x) { if (x < 0) {x = -x;putchar ('-');}if (x > 9) write (x / 10);putchar (x % 10 + '0'); }
template <typename T> void print (T x, char ch) { write (x); putchar (ch); }
template <typename T> T Max (T x, T y) { return x > y ? x : y; }
template <typename T> T Min (T x, T y) { return x < y ? x : y; }
template <typename T> T Abs (T x) { return x > 0 ? x : -x; }
const int Maxe = 1e5;
int Num_E;
struct Equation {
LL Mod, b;
}Eq[Maxe + 5];
LL md (LL x, LL MO) {
return (x % MO + MO) % MO;
}
LL gcd (LL x, LL y) {
if (y == 0) return x;
else return gcd (y, x % y);
}
LL lcm (LL x, LL y) {
return x / gcd (x, y) * y;
}
LL mul (LL x, LL y, LL Mod) {
LL res = 0;
while (y) {
if (y & 1) res = (res + x) % Mod;
x = (x + x) % Mod; y >>= 1;
}
return res;
}
void exgcd (LL a, LL b, LL &x, LL &y) {
if (b == 0) {
x = 1; y = 0;
return ;
}
exgcd (b, a % b, y, x);
y -= (a / b) * x;
}
LL solve (LL a, LL b, LL c) {
LL _gcd = gcd (a, b);
c = (c % b + b) % b;
if (c % _gcd != 0) return -1;
LL x, y;
exgcd (a, b, x, y);
LL p = b / _gcd;
x = (x % p + p) % p;
x = mul (x, (c / _gcd), p);
return x;
}
LL quick_pow (LL x, LL y, LL Mod) {
LL res = 1;
while (y) {
if (y & 1) res = mul (res, x, Mod);
x = mul (x, x, Mod); y >>= 1;
}
return res;
}
PLL Excrt () {//求同余方程组的模板
LL Mod = 1, b = 0;
for (int i = 1; i <= Num_E; i++) {
LL res = solve (Mod, Eq[i].Mod, Eq[i].b - b);
if (res == -1) {
return MP (-1, -1);
}
LL M = lcm (Mod, Eq[i].Mod);
b = (b + res * Mod % M) % M;
Mod = M;
}
return MP (b, Mod);
}
const int Maxn = 1e5;
int t, n, m;
LL a[Maxn + 5], p[Maxn + 5], sword[Maxn + 5];
multiset <LL> atk;
int main () {
// freopen ("dragon.in", "r", stdin);
// freopen ("dragon.out", "w", stdout);
read (t);
while (t--) {
atk.clear ();
read (n); read (m);
for (int i = 1; i <= n; i++) read (a[i]);
for (int i = 1; i <= n; i++) read (p[i]);
for (int i = 1; i <= n; i++) read (sword[i]);
for (int i = 1; i <= m; i++) {
int x; read (x);
atk.insert (x);
}
Num_E = n;
bool flag = 0; LL _max = 0;
for (int i = 1; i <= n; i++) {
//这里我用的是 multiset 找到对应的剑
auto it = atk.upper_bound (a[i]);
if (it != atk.begin ()) it--;
_max = Max (_max, (LL)ceil (a[i] * 1.0 / *it));
//找到至少要攻击多少次
LL x, y;
LL _gcd = gcd (p[i], *it);
if (a[i] % _gcd != 0) {
//裴蜀定理，没有满足要求的解
flag = 1;
break;
}
exgcd (*it, p[i], x, y);
LL Mod = p[i] / _gcd;
x = md (x, p[i]);
//x ： 求出一个特解
//Mod ： 求出周期
Eq[i].b = mul (x, a[i] / _gcd, p[i]);
Eq[i].Mod = Mod;
atk.erase (it);
atk.insert (sword[i]);
}
if (flag == 1) {
print (-1, '\n');
continue;
}
PLL res = Excrt ();
//res = (特解，模数的 lcm (周期))
if (res.first == -1) {
print (-1, '\n');
continue;
}
res.first += ceil ((_max - res.first) * 1.0 / res.second) * res.second;
//修正答案
print (res.first, '\n');
}
return 0;
}
