辗转相减法的扩展 $gcd(x, y, z) = gcd(x, y - x, z - y)$ 当有n个数时也成立
所以构造$a_{i}$的差分数组$b_{i} = a_{i} - a_{i - 1}$，用一个线段树来维护b数组的gcd，这样每次区间修改相当于两次单点修改
考虑到询问的时候$ans = gcd(a_{l}, query(l +1, r))$所以我们再维护原数组a的值，直接差分之后用一个树状数组就好了
注意判断边界情况。
Code：
#include <cstdio> #include <cstring> using namespace std; typedef long long ll;const int N = 5e5 + 5;int n, qn; ll a[N], b[N];ll gcd(ll x, ll y) {return (!y) ? x : gcd(y, x % y); }template <typename T> inline void read(T &X) {X = 0;char ch = 0;T op = 1;for(; ch > '9' || ch < '0'; ch = getchar())if(ch == '-') op = -1;for(; ch >= '0' && ch <= '9'; ch = getchar())X = (X << 3) + (X << 1) + ch - 48;X *= op; }struct BinaryIndexTree {ll s[N];#define lowbit(x) ((x) & (-x))inline void add(int x, ll v) {for(; x <= n; x += lowbit(x))s[x] += v;} inline ll query(int x) {ll res = 0;for(; x > 0; x -= lowbit(x))res += s[x];return res;}} B;struct SegT {ll s[N << 2];#define lc p << 1#define rc p << 1 | 1#define mid ((l + r) >> 1)inline void up(int p) {if(p) s[p] = gcd(s[lc], s[rc]);}void build(int p, int l, int r) {if(l == r) {s[p] = b[l];return;}build(lc, l, mid);build(rc, mid + 1, r);up(p);}void modify(int p, int l, int r, int x, ll v) {if(x == l && x == r) {s[p] += v;return;}if(x <= mid) modify(lc, l, mid, x, v);else modify(rc, mid + 1, r, x, v);up(p);}ll query(int p, int l, int r, int x, int y) {if(x > y) return 1LL;if(x <= l && y >= r) return s[p];ll res;if(x <= mid && y > mid) res = gcd(query(lc, l, mid, x, y), query(rc, mid + 1, r, x, y)); else if(y <= mid) res = query(lc, l, mid, x, y);else if(x > mid) res = query(rc, mid + 1, r, x, y);return res;}} A; inline ll abs(ll x) {return x > 0 ? x : -x; }int main() {read(n), read(qn);for(int i = 1; i <= n; i++) read(a[i]);for(int i = 1; i <= n; i++) b[i] = a[i] - a[i - 1];for(int i = 1; i <= n; i++) B.add(i, b[i]);A.build(1, 1, n);char op[3];for(int x, y; qn--; ) {scanf("%s", op);read(x), read(y);if(op[0] == 'C') {ll v;read(v);B.add(x, v), A.modify(1, 1, n, x, v);if(y < n) B.add(y + 1, -v), A.modify(1, 1, n, y + 1, -v);} else {if(x < y) printf("%lld\n", gcd(B.query(x), abs(A.query(1, 1, n, x + 1, y))));else printf("%lld\n", B.query(x));} }return 0; }