其他分享
首页 > 其他分享> > Gym 237040F 线段树:区间修改,区间查询

Gym 237040F 线段树:区间修改,区间查询

作者:互联网

题目链接:Gym 237040F 线段树:区间修改,区间查询

题目大意:

题解:

线段树或者树状数组模板题。
\(cin,cout\)貌似会超时,要用\(scanf,printf\)或者快读。


线段树

考察对\(lazy\)标记的使用。

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
#define ll long long

struct Tree {
    ll num, left, right, lazy;
} tree[4000005];

ll read() {
    char x = getchar();
    ll ans = 0, f = 1;
    while (!isdigit(x)) {
        if (x == '-') f = -1;
        x = getchar();
    }
    while (isdigit(x)) {
        ans = (ans << 3) + (ans << 1) + (x ^ 48);
        x = getchar();
    }
    return ans * f;
}

void build(ll k, ll l, ll r) {
    tree[k].lazy = 0, tree[k].left = l, tree[k].right = r;
    if (l == r) {
        tree[k].num = read();
        return;
    }
    ll mid = l + r >> 1;
    build(k << 1, l, mid);
    build((k << 1) | 1, mid + 1, r);
    tree[k].num = tree[k << 1].num + tree[(k << 1) | 1].num;
}

void push_down(ll k) {
    tree[k << 1].lazy += tree[k].lazy;
    tree[(k << 1) | 1].lazy += tree[k].lazy;
    tree[k << 1].num += tree[k].lazy * (tree[k << 1].right - tree[k << 1].left + 1);
    tree[(k << 1) | 1].num += tree[k].lazy * (tree[(k << 1) | 1].right - tree[(k << 1) | 1].left + 1);
    tree[k].lazy = 0;
}

void update(ll k, ll l, ll r, ll add) {
    if (tree[k].left >= l && tree[k].right <= r) {
        tree[k].num += add * (tree[k].right - tree[k].left + 1);
        tree[k].lazy += add;
        return;
    }
    if (tree[k].lazy) push_down(k);
    ll mid = tree[k].left + tree[k].right >> 1;
    if (l <= mid) update(k << 1, l, r, add);
    if (r > mid) update((k << 1) | 1, l, r, add);
    tree[k].num = tree[k << 1].num + tree[(k << 1) | 1].num;
}

ll query(ll k, ll l, ll r) {
    if (tree[k].left >= l && tree[k].right <= r) {
        return tree[k].num;
    }
    if (tree[k].lazy) push_down(k);
    ll mid = tree[k].left + tree[k].right >> 1;
    ll ans = 0;
    if (l <= mid) ans += query(k << 1, l, r);
    if (r > mid) ans += query((k << 1) | 1, l, r);
    return ans;
}

int main() {
    ll n, m, add, x, y, act;
    n = read(), m = read();
    build(1, 1, n);
    while (m--) {
        act = read(), x = read(), y = read();
        if (act == 1) {
            add = read();
            update(1, x, y, add);
        }
        if (act == 2) {
            printf("%lld\n", query(1, x, y));
        }
    }
    return 0;
}

树状数组

#include <cstdio>
#include <iostream>
using namespace std;

long long n, q, bit0[1000010], bit1[1000010];

void updata(long long *bit, long long x, long long add) {
    while (x <= n) {
        bit[x] += add;
        x += x & -x;
    }
}

long long getSum(long long *bit, long long x) {
    long long sum = 0;
    while (x > 0) {
        sum += bit[x];
        x -= x & -x;
    }
    return sum;
}

int main() {
    scanf("%lld%lld", &n, &q);
    long long act, add, l, r;
    for (int i = 1; i <= n; ++i) {
        scanf("%lld", &add);
        updata(bit0, i, add);
    }
    for (int i = 1; i <= q; ++i) {
        scanf("%lld", &act);
        if (act == 1) {
            scanf("%lld%lld%lld", &l, &r, &add);
            updata(bit1, l, add);
            updata(bit0, l, -add * (l - 1));
            updata(bit1, r + 1, -add);
            updata(bit0, r + 1, add * r);
        } else {
            scanf("%lld%lld", &l, &r);
            long long ans = 0;
            ans += getSum(bit1, r) * r + getSum(bit0, r);
            ans -= getSum(bit1, l - 1) * (l - 1) + getSum(bit0, l - 1);
            printf("%lld\n", ans);
        }
    }
    return 0;
}

标签:线段,right,Gym,237040F,long,ans,区间,include,ll
来源: https://www.cnblogs.com/IzumiSagiri/p/15860596.html