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 + ; 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 = ;
char ch = ;
T op = ;
for(; ch > '' || ch < ''; ch = getchar())
if(ch == '-') op = -;
for(; ch >= '' && ch <= ''; ch = getchar())
X = (X << ) + (X << ) + ch - ;
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 = ;
for(; x > ; x -= lowbit(x))
res += s[x];
return res;
} } B; struct SegT {
ll s[N << ]; #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 + , 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 + , 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 + , r, x, y));
else if(y <= mid) res = query(lc, l, mid, x, y);
else if(x > mid) res = query(rc, mid + , r, x, y);
return res;
} } A; inline ll abs(ll x) {
return x > ? x : -x;
} int main() {
read(n), read(qn);
for(int i = ; i <= n; i++) read(a[i]); for(int i = ; i <= n; i++) b[i] = a[i] - a[i - ];
for(int i = ; i <= n; i++) B.add(i, b[i]);
A.build(, , n); char op[];
for(int x, y; qn--; ) {
scanf("%s", op);
read(x), read(y);
if(op[] == 'C') {
ll v;
read(v);
B.add(x, v), A.modify(, , n, x, v);
if(y < n) B.add(y + , -v), A.modify(, , n, y + , -v);
} else {
if(x < y) printf("%lld\n", gcd(B.query(x), abs(A.query(, , n, x + , y))));
else printf("%lld\n", B.query(x));
}
} return ;
}