CH 4302 Interval GCD

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
", gcd(B.query(x), abs(A.query(1, 1, n, x + 1, y))));
            else printf("%lld
", B.query(x));
        }        
    }
    
    return 0;
}