\[\text{orz lxl sto} \]
Codechef DGCD (Weaker) / AcWing 246
给定一个长度为 \(n\) 的数列 \(A = (a_1, a_2, \dots, a_n)\)，支持两种操作：
-
C L R d：将 \(a_L, a_{L+1}, \dots, a_R\) 都加上 \(d\)。 -
Q L R：查询 \(\gcd(a_L, a_{L+1}, \dots, a_R)\)。
\(1 \leq n \leq 500000\)。
根据辗转相减法，有：
\[\gcd(x,y)=\gcd(x,y-x) \]拓展一下，就有：
\[\gcd(p_1, p_2, \dots, p_m) = \gcd(p_1, p_2 - p_1, p_3 - p_2, \dots, p_m - p_{m-1}) \]如果加上 \(d\)，影响是：
\[\gcd(p_1 + d, p_2 + d, \dots, p_m + d) = \gcd(p_1 + d, p_2 - p_1, p_3 - p_2, \dots, p_m - p_{m-1}) \]维护两棵线段树即可。（因为单纯的区间加线段树会爆 long long，所以我们采取区间 min 的线段树，反正也只需要查询单点值）
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/hash_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
#ifdef LOCAL
#include "algo/debug.h"
#else
#define debug(...) 42
#endif
typedef long long ll;
typedef pair < int, int > PII;
typedef int itn;
mt19937 RND_MAKER (chrono :: steady_clock :: now ().time_since_epoch ().count ());
inline ll randomly (const ll l, const ll r) {return (RND_MAKER () ^ (1ull << 63)) % (r - l + 1) + l;}
#define int long long
const double pi = acos (-1);
//__gnu_pbds :: tree < Key, Mapped, Cmp_Fn = std :: less < Key >, Tag = rb_tree_tag, Node_Upadte = null_tree_node_update, Allocator = std :: allocator < char > > ;
//__gnu_pbds :: tree < PPS, __gnu_pbds :: null_type, less < PPS >, __gnu_pbds :: rb_tree_tag, __gnu_pbds :: tree_order_statistics_node_update > tr;
inline int read () {
int x = 0, f = 0;
char c = getchar ();
for ( ; c < '0' || c > '9' ; c = getchar ()) f |= (c == '-');
for ( ; c >= '0' && c <= '9' ; c = getchar ()) x = (x << 1) + (x << 3) + (c & 15);
return !f ? x : -x;
}
const int N = 5e5 + 5;
namespace ori {
int mn[N << 2], inc[N << 2], seq[N];
inline void push_up (int u) {mn[u] = min (mn[u << 1], mn[u << 1 | 1]);}
inline void push_down (int u) {
if (!inc[u]) return ;
inc[u << 1] += inc[u], inc[u << 1 | 1] += inc[u];
mn[u << 1] += inc[u], mn[u << 1 | 1] += inc[u];
inc[u] = 0;
}
inline void build (int u, int l, int r) {
inc[u] = 0;
if (l == r) {mn[u] = seq[l]; return ;}
int mid = l + r >> 1;
build (u << 1, l, mid), build (u << 1 | 1, mid + 1, r);
push_up (u);
}
inline void modify (int u, int l, int r, int x, int y, int v) {
if (x <= l && r <= y) {mn[u] += v, inc[u] += v; return ;}
push_down (u);
int mid = l + r >> 1;
if (x <= mid) modify (u << 1, l, mid, x, y, v);
if (y > mid) modify (u << 1 | 1, mid + 1, r, x, y, v);
push_up (u);
}
inline int query (int u, int l, int r, int x, int y) {
if (x <= l && r <= y) return mn[u];
push_down (u);
int mid = l + r >> 1, ans = 9223372036854775807ll;
if (x <= mid) ans = min (ans, query (u << 1, l, mid, x, y));
if (y > mid) ans = min (ans, query (u << 1 | 1, mid + 1, r, x, y));
return ans;
}
}
inline int gcd (int a, int b) {
if (!a || !b) return a + b;
else {
int tag = ((a < 0) && (b < 0));
a = abs (a), b = abs (b);
if (tag) tag = -1; else tag = 1;
return tag * gcd (b, a % b);
}
}
namespace diff {
int val[N << 2], inc[N << 2], seq[N];
inline void push_up (int u) {val[u] = gcd (val[u << 1], val[u << 1 | 1]);}
inline void push_down (int u) {
if (inc[u] == 9223372036854775807ll) return ;
inc[u << 1] = inc[u << 1 | 1] = val[u << 1] = val[u << 1 | 1] = inc[u];
inc[u] = 9223372036854775807ll;
}
inline void build (int u, int l, int r) {
inc[u] = 9223372036854775807ll;
if (l == r) {val[u] = seq[l]; return ;}
int mid = l + r >> 1;
build (u << 1, l, mid), build (u << 1 | 1, mid + 1, r);
push_up (u);
}
inline void update (int u, int l, int r, int x, int y, int v) {
if (x <= l && r <= y) {val[u] = inc[u] = v; return ;}
push_down (u);
int mid = l + r >> 1;
if (x <= mid) update (u << 1, l, mid, x, y, v);
if (y > mid) update (u << 1 | 1, mid + 1, r, x, y, v);
push_up (u);
}
inline int query (int u, int l, int r, int x, int y) {
if (x <= l && r <= y) return val[u];
int mid = l + r >> 1, ans = 0;
if (x <= mid) ans = gcd (ans, query (u << 1, l, mid, x, y));
if (y > mid) ans = gcd (ans, query (u << 1 | 1, mid + 1, r, x, y));
return ans;
}
}
int n, m;
signed main () {
n = read (), m = read ();
ori :: seq[0] = 0;
for (int i = 1;i <= n; ++ i) ori :: seq[i] = read ();
for (int i = 1;i <= n; ++ i) diff :: seq[i] = ori :: seq[i] - ori :: seq[i - 1];
ori :: build (1, 1, n), diff :: build (1, 1, n);
while (m --) {
char op[2];
scanf ("%s", op + 1);
if (op[1] == 'C') {
int l = read (), r = read (), d = read ();
ori :: modify (1, 1, n, l, r, d);
int val1 = ori :: query (1, 1, n, l, l);
if (l > 1) val1 -= ori :: query (1, 1, n, l - 1, l - 1);
diff :: update (1, 1, n, l, l, val1);
if (r + 1 <= n) diff :: update (1, 1, n, r + 1, r + 1, ori :: query (1, 1, n, r + 1, r + 1) - ori :: query (1, 1, n, r, r));
}
else {
int l = read (), r = read ();
int res = ori :: query (1, 1, n, l, l);
if (l + 1 <= r) res = gcd (res, diff :: query (1, 1, n, l + 1, r));
printf ("%lld\n", res);
}
}
return 0;
}