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  • AcWing 246. 区间最大公约数

    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 << "
    ";
    #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
    ", __gcd(a[l]+B.sum(l), query(l+1, r, 1, 1, n)));
            }
        }
        return 0;
    }
    
    
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  • 原文地址:https://www.cnblogs.com/widsom/p/11259953.html
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