AcWing 246. 区间最大公约数

Posted 2020-11-10 Wisdom+.+

246. 区间最大公约数
思路：
首先根据更相减损术，我们得到一个结论：
\(gcd(a_l, a_{l+1}, ...,a_r) = gcd(a_l, a_{l+1}-a_l, a_{l+2}-a_{l+1}, ..., a_r-a_{r-1})\)
于是我们用线段树维护差分数组，树状数组维护每个位置的值，然后查询就是\(gcd(a_l+bit.sum(l), segtree.query(l+1, r))\)。
代码：

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
#include<bits/stdc++.h>
using namespace std;
#define y1 y11
#define fi first
#define se second
#define pi acos(-1.0)
#define LL long long
//#define mp make_pair
#define pb emplace_back
#define ls rt<<1, l, m
#define rs rt<<1|1, m+1, r
#define ULL unsigned LL
#define pll pair<LL, LL>
#define pli pair<LL, int>
#define pii pair<int, int>
#define piii pair<pii, int>
#define pdd pair<double, double>
#define mem(a, b) memset(a, b, sizeof(a))
#define debug(x) cerr << #x << " = " << x << "\n";
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
//head
const int N = 5e5 + 5;
int n, m, l, r;
LL a[N], d;
char op[15];
LL tree[N<<2];
struct BIT {
    LL bit[N];
    void add(int x, LL a) {
        while(x <= n) bit[x] += a, x += x&-x;
    }
    LL sum(int x) {
        LL res = 0;
        while(x) res += bit[x], x -= x&-x;
        return res;
    }
    void init() {
        for (int i = 1; i <= n; ++i) bit[i] = 0;
    }
}B;
inline void push_up(int rt) {
    tree[rt] = __gcd(tree[rt<<1], tree[rt<<1|1]);
}
LL query(int L, int R, int rt, int l, int r) {
    if(L > R) return 0;
    if(L <= l && r <= R) return tree[rt];
    int m = l+r >> 1;
    LL res = 0;
    if(L <= m) res = __gcd(res, query(L, R, ls));
    if(R > m) res = __gcd(res, query(L, R, rs));
    return abs(res);
}
void update(int p, LL d, int rt, int l, int r) {
    if(p > r) return ;
    if(l == r) {
        tree[rt] += d;
        return ;
    }
    int m = l+r >> 1;
    if(p <= m) update(p, d, ls);
    else update(p, d, rs);
    push_up(rt);
}
void build(int rt, int l, int r) {
    if(l == r) {
        tree[rt] = a[l]-a[l-1];
        return ;
    }
    int m = l+r >> 1;
    build(ls);
    build(rs);
    push_up(rt);
}
int main() {
    scanf("%d %d", &n, &m);
    for (int i = 1; i <= n; ++i) scanf("%lld", &a[i]);
    build(1, 1, n);
    B.init();
    while(m--) {
        scanf("%s", op);
        if(op[0] == 'C') {
            scanf("%d %d %lld", &l, &r, &d);
            B.add(l, d);
            update(l, d, 1, 1, n);
            B.add(r+1, -d);
            update(r+1, -d, 1, 1, n);
        }
        else {
            scanf("%d %d", &l, &r);
            printf("%lld\n", __gcd(a[l]+B.sum(l), query(l+1, r, 1, 1, n)));
        }
    }
    return 0;
}