poj 3468 A Simple Problem with Integers(线段树)

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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;}

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