poj 3468 A Simple Problem with Integers(线段树)

题目链接：poj 3468 A Simple Problem with Integers

题目大意：给定一个序列，有Q次操作，询问一段区间的总和，为一段区间的元素统一加上v值。

解题思路：线段树的成段修改。

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

#define lson(x) ((x)<<1)
#define rson(x) (((x)<<1)+1)

const int maxn = 100005;
typedef long long ll;

struct Node {
    int l, r;
    ll sum, add;
    void set (int l, int r, ll sum, ll add) {
        this->l = l;
        this->r = r;
        this->sum = sum;
        this->add = add;
    }
    void maintain (ll v) {
        sum += (r - l + 1) * v;
        add += v;
    }
}nd[maxn * 4];

int N, Q, A[maxn];

void pushup (int u) {
    nd[u].sum = nd[lson(u)].sum + nd[rson(u)].sum;
}

void pushdown(int u) {
    if (nd[u].add) {
        nd[lson(u)].maintain(nd[u].add);
        nd[rson(u)].maintain(nd[u].add);
        nd[u].add = 0;
    }
}

void build (int u, int l, int r) {
    nd[u].l = l;
    nd[u].r = r;
    nd[u].add = 0;

    if (l == r) {
        nd[u].sum = A[l];
        return;
    }

    int mid = (l + r) / 2;
    build(lson(u), l, mid);
    build(rson(u), mid + 1, r);
    pushup(u);
}

ll query(int u, int l, int r) {
    if (l <= nd[u].l && nd[u].r <= r)
        return nd[u].sum;

    pushdown(u);
    ll ret = 0;
    int mid = (nd[u].l + nd[u].r) / 2;
    if (l <= mid)
        ret += query(lson(u), l, r);
    if (r > mid)
        ret += query(rson(u), l, r);
    pushup(u);
    return ret;
}

void modify (int u, int l, int r, int v) {
    if (l <= nd[u].l && nd[u].r <= r) {
        nd[u].maintain(v);
        return;
    }

    pushdown(u);
    int mid = (nd[u].l + nd[u].r) / 2;
    if (l <= mid)
        modify(lson(u), l, r, v);
    if (r > mid)
        modify(rson(u), l, r, v);
    pushup(u);
}

int main () {
    while (scanf("%d%d", &N, &Q) == 2) {
        for (int i = 1; i <= N; i++)
            scanf("%d", &A[i]);

        build(1, 1, N);

        int l, r, v;
        char order[5];
        for (int i = 0; i < Q; i++) {
            scanf("%s%d%d", order, &l, &r);
            if (order[0] == 'Q')
                printf("%lld\n", query(1, l, r));
            else {
                scanf("%d", &v);
                modify(1, l, r, v);
            }
        }
    }
    return 0;
}