题意
给定数组(a)和(b)，长度分别为(n)和(m)。对于每一个(b_j),输出(GCD(a_1 + b_j, a_2 + b_j, ..., a_n + b_j))。
解法
经典区间GCD。
首先根据GCD的性质，可以得出(GCD(a, b, c) = GCD(a, b - a, c - b))。这个结论可以推广到n维。
所以就可以维护(a)的差分，计算出差分数组中(n - 1)个元素的GCD, 即(g = GCD(g, a_i - a_{i - 1}) forall i ge 2)，然后对于(b_j)，答案就是(GCD(a_1 + b_j, g))。
注意，因为差分数组中可能存在负数，导致最后的(g)维负数，所以在输出时记得取绝对值。
AC代码
#include <bits/stdc++.h>
using namespace std;
using ll = int64_t;
using ull = uint64_t;
using VI = vector<int>;
using VL = vector<ll>;
using VVI = vector<vector<int>>;
using VVL = vector<vector<ll>>;
using PII = pair<int,int>;
using PLL = pair<ll, ll>;
#define REP(i, _, __) for (int i = (_); i < (__); ++i)
#define PER(i, _, __) for (int i = (_-1); i >= (__); --i)
#define FOR(i, _, __) for (int i = (_); i <= (__); ++i)
#define ROF(i, _, __) for (int i = (_); i >= (__); --i)
#define FE(v, V) for (const auto& v: V)
#define EB emplace_back
#define PB push_back
#define MP make_pair
#define FI first
#define SE second
#define SZ(x) (int((x).size()))
#define ALL(x) (x).begin(),(x).end()
#define LLA(x) (x).rbegin(),(x).rend()
const double PI = acos(-1.0);
namespace Backlight {
const int __BUFFER_SIZE__ = 1 << 20;
bool NEOF = 1;
int __top;
char __buf[__BUFFER_SIZE__], *__p1 = __buf, *__p2 = __buf, __stk[996];
template<typename T>
T MIN(T a, T b) { return min(a, b); }
template<typename First, typename... Rest>
First MIN(First f, Rest... r) { return min(f, MIN(r...)); }
template<typename T>
T MAX(T a, T b) { return max(a, b); }
template<typename First, typename... Rest>
First MAX(First f, Rest... r) { return max(f, MAX(r...)); }
template<typename T>
void updMin(T& a, T b) { if (a > b) a = b; }
template<typename T>
void updMax(T& a, T b) { if (a < b) a = b; }
inline char nc() {
return __p1 == __p2 && NEOF && (__p2 = (__p1 = __buf) + fread(__buf, 1, __BUFFER_SIZE__, stdin), __p1 == __p2) ? (NEOF = 0, EOF) : *__p1++;
}
template<typename T>
inline bool read(T &x) {
char c = nc();
bool f = 0; x = 0;
while (!isdigit(c)) c == '-' && (f = 1), c = nc();
while (isdigit(c)) x = (x << 3) + (x << 1) + (c ^ 48), c = nc();
if (f) x = -x;
return NEOF;
}
inline bool need(char c) { return (c != '
') && (c != ' '); }
inline bool read(char& a) {
while ((a = nc()) && need(a) && NEOF) ;
return NEOF;
}
inline bool read(char *a) {
while ((*a = nc()) && need(*a) && NEOF) ++a;
*a = ' ';
return NEOF;
}
inline bool read(double &x) {
bool f = 0; char c = nc(); x = 0;
while (!isdigit(c)) { f |= (c == '-'); c = nc(); }
while (isdigit(c)) { x = x * 10.0 + (c ^ 48); c = nc(); }
if (c == '.') {
double temp = 1; c = nc();
while (isdigit(c)) { temp = temp / 10.0; x = x + temp * (c ^ 48); c = nc(); }
}
if (f) x = -x;
return NEOF;
}
template<typename T, typename... T2>
inline bool read(T &x, T2 &... rest) {
read(x);
return read(rest...);
}
template<typename T>
inline void print(T x) {
if (x < 0) putchar('-'), x = -x;
if (x == 0) { putchar('0'); return; }
__top = 0;
while(x) {
__stk[++__top] = x % 10 + '0';
x /= 10;
}
while(__top) {
putchar(__stk[__top]);
--__top;
}
}
template<typename First, typename... Rest>
inline void print(First f, Rest... r) {
print(f); putchar(' ');
print(r...);
}
template<typename T>
inline void println(T x) {
print(x);
putchar('
');
}
template<typename First, typename... Rest>
inline void println(First f, Rest... r) {
print(f); putchar(' ');
println(r...);
}
template<typename T>
inline void _dbg(const char *format, T value) { cerr << format << '=' << value << endl; }
template<typename First, typename... Rest>
inline void _dbg(const char *format, First f, Rest... r) {
while(*format != ',') cerr << *format++;
cerr << '=' << f << ", ";
_dbg(format + 1, r...);
}
template<typename T>
ostream &operator<<(ostream& os, vector<T> V) {
os << "[ "; for (auto v : V) os << v << ","; return os << " ]";
}
template<typename T>
ostream &operator<<(ostream& os, set<T> V) {
os << "[ "; for (auto v : V) os << v << ","; return os << " ]";
}
template<typename T>
ostream &operator<<(ostream& os, multiset<T> V) {
os << "[ "; for (auto v : V) os << v << ","; return os << " ]";
}
template<typename T1, typename T2>
ostream &operator<<(ostream& os, map<T1, T2> V) {
os << "[ "; for (auto v : V) os << v << ","; return os << " ]";
}
template<typename L, typename R>
ostream &operator<<(ostream &os, pair<L, R> P) {
return os << "(" << P.first << "," << P.second << ")";
}
#ifdef BACKLIGHT
#define debug(...) cerr << " 33[31m" << "[" << __LINE__ << "] : "; _dbg(#__VA_ARGS__, __VA_ARGS__); cerr << " 33[0m";
// #define debug(...) _dbg(#__VA_ARGS__, __VA_ARGS__);
#else
#define debug(...)
#endif
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int rnd(int l, int r) { return l + rng() % (r - l + 1); }
using namespace Backlight;
const int N = 5e5 + 5;
const int M = 5e5 + 5;
const int K = 1e7 + 5;
const int MOD = 1e9 + 7; // 998244353 1e9 + 7
const int INF = 0x3f3f3f3f; // 1e9 + 7 0x3f3f3f3f
const ll LLINF = 0x3f3f3f3f3f3f3f3f; // 1e18 + 9 0x3f3f3f3f3f3f3f3f
const double eps = 1e-8;
ll n, m, a[N], b[N], d[N];
void solve(int Case) { // printf("Case #%d: ", Case);
read(n, m);
FOR(i, 1, n) read(a[i]);
FOR(i, 1, m) read(b[i]);
FOR(i, 1, n) d[i] = a[i + 1] - a[i];
ll g = 0;
FOR(i, 1, n - 1) g = __gcd(g, d[i]);
FOR(i, 1, m) printf("%lld ", abs(__gcd(g, a[1] + b[i])));
puts("");
}
int main() {
#ifdef BACKLIGHT
freopen("in.txt", "r", stdin);
// freopen("out.txt", "w", stdout);
auto begin = std::chrono::steady_clock::now();
#endif
// ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
int T = 1;
// read(T);
for (int _ = 1; _ <= T; _++) solve(_);
#ifdef BACKLIGHT
auto end = std::chrono::steady_clock::now();
auto duration = std::chrono::duration_cast<std::chrono::milliseconds>(end - begin);
cerr << " 33[32mTime Elasped: " << duration.count() << " ms 33[0m" << endl;
#endif
return 0;
}