POJ - 3468 　A Simple Problem with Integers

最新推荐文章于 2022-08-28 01:34:07 发布

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本文介绍了一种使用线段树进行区间更新和查询的高效算法，通过预处理和延迟传播实现快速操作。适用于解决诸如区间求和、区间修改等动态数组问题，常见于ACM竞赛和在线评测平台。

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题目链接：A Simple Problem with Integers

#include <iostream>
#include <algorithm>
#include <map>
#include <stack>
#include <queue>
#include <set>
#include <vector>
#include <cstdlib>
#include <cmath>
#include <string>
#include <cstdio>
#include <cstring>
#include <assert.h>

#define  lowbit(x) (x&(-x))
#define  ll long long
#define  ull unsigned long long
#define  max(a, b) (a>b?a:b)
#define  min(a, b) (a>b?b:a)
#define  PI acos(-1.0)
#define  fir first
#define  sec second
#define  SZ(x) (x.size())
#define  inf 0x3f3f3f3f
#define  PB push_back
#define  MP make_pair
#define  all(x) (x).begin(),(x).end()
#define  PII pair<int,int>
#define  E 2.7182818284590452353602874713527
using namespace std;

#define  ll long long
#define all(x) (x).begin(),(x).end()
using namespace std;

void scan() {
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif
}

const int INF = 0x3f3f3f3f;
const int maxn = 2e5 + 7;
ll a[maxn], st[maxn << 2], cha[maxn << 2];


void build(int o, int l, int r) {
    if (l == r) {
        st[o] = a[l];
        return;
    }
    int mid = l + ((r - l >> 1));
    build(o << 1, l, mid);
    build(o << 1 | 1, mid + 1, r);
    st[o] = st[o << 1] + st[o << 1 | 1];
}


void push_up(int o) {
    st[o] = st[o << 1] + st[o << 1 | 1];
}

void push_down(int o, int l, int r) {
    if (cha[o]) {
        cha[o << 1] += cha[o];
        cha[o << 1 | 1] += cha[o];
        int mid = l + ((r - l) >> 1);
        st[o << 1] += cha[o] * (mid - l + 1);
        st[o << 1 | 1] += cha[o] * (r - mid);
        cha[o] = 0;
    }
}


void update(int o, int l, int r, int ql, int qr, ll c) {//区间加
    if (ql <= l && qr >= r) {
        cha[o] += c;
        st[o] += 1ll * (r - l + 1) * c;
        return;
    }

    push_down(o, l, r);
    int mid = l + ((r - l) >> 1);
    if (ql <= mid) update(o << 1, l, mid, ql, qr, c);
    if (qr >= mid + 1) update(o << 1 | 1, mid + 1, r, ql, qr, c);
    push_up(o);
}


ll query(int o, int l, int r, int ql, int qr) {//和
    if (ql <= l && qr >= r) return st[o];
    int mid = l + ((r - l) >> 1);
    push_down(o, l, r);
    ll ans = 0;
    if (ql <= mid) ans += query(o << 1, l, mid, ql, qr);
    if (qr >= mid + 1) ans += query(o << 1 | 1, mid + 1, r, ql, qr);
    return ans;
}


int main() {
    int n;
    scanf("%d", &n);
    int Q;
    scanf("%d", &Q);
    for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
    memset(cha, 0, sizeof(cha));
    build(1, 1, n);
    while (Q--) {
        string ch;
        cin >> ch;
        int ql, qr;
        scanf("%d %d", &ql, &qr);
        if (ch[0] == 'Q') printf("%lld\n", query(1, 1, n, ql, qr));
        else {
            ll c;
            scanf("%lld", &c);
            update(1, 1, n, ql, qr, c);
        }
    }
    return 0;
}