线段树 ：区间修改，区间查询

题目描述

思路

代码

#include <cstdio>
#include <cstring>
long long n, m;
struct {
    long long at[1000005], arr[1000005 << 2], ad[1000005 << 2];
    void build(int k, int l, int r) {
        if (l == r) {
            arr[k] = at[l];
            return;
        }
        int mid = l + r >> 1;
        this->build(k << 1, l, mid);
        this->build(k << 1 | 1, mid + 1, r);
        arr[k] = arr[k << 1] + arr[k << 1 | 1];
    }
    void change(int k, int l, int r, int x, int y, int v) {
        if (x <= l && r <= y) {
            this->add(k, l, r, v);
            return;
        }
        int mid = l + r >> 1;
        this->pushdown(k, l, r, mid);
        if (x <= mid) this->change(k << 1, l, mid, x, y, v);
        if (y > mid) this->change(k << 1 | 1, mid + 1, r, x, y, v);
        arr[k] = arr[k << 1] + arr[k << 1 | 1]; 
    }
    long long query(int k, int l, int r, int x, int y) {
        if (x <= l && r <= y) return arr[k];
        long long res = 0;
        int mid = l + r >> 1;
        this->pushdown(k, l, r, mid);
        if (x <= mid) res += this->query(k << 1, l, mid, x, y);
        if (y > mid) res += this->query(k << 1 | 1, mid + 1, r, x, y);
        return res;
    }
    void add(int k, int l, int r, int v) {
        ad[k] += v;
        arr[k] += (long long)(r - l + 1) * v;
    }
    void pushdown(int k, int l, int r, int mid) {
        if (ad[k] == 0) return;
        add(k << 1, l, mid, ad[k]);
        add(k << 1 | 1, mid + 1, r, ad[k]);
        ad[k] = 0;
    }
} t;

inline long long read() {
    long long s = 0, f = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9') {
        if (ch == '-') f = -1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9') s = s * 10 + ch - '0', ch = getchar();
    return s * f;
}
int main() {
    n = read(), m = read();
    for (register int i = 1; i <= n; ++i) t.at[i] = read();
    t.build(1, 1, n);
    for (register int i = 1, a, b, c, d; i <= m; ++i) {
        a = read();
        if (a == 1) {
            b = read(), c = read(), d = read();
            t.change(1, 1, n, b, c, d);
        } else {
            b = read(), c = read();
            printf("%lld\n", t.query(1, 1, n, b, c));
        }
    }
    return 0;
}

原文地址：https://www.cnblogs.com/liuzz-20180701/p/11494578.html