POJ3468-A Simple Problem with Integers

一道经典的线段树区间更新的题，你值得拥有！

#include <cstdio>#define lchild rt << 1, l, m#define rchild rt << 1 | 1, m + 1, rconst int maxn = 1 << 18;long long tree[maxn];long long lazy[maxn];int n;void push_up(int rt) {    tree[rt] = tree[rt<<1] + tree[rt<<1|1];}void push_down(int rt, int len) {    tree[rt<<1] += lazy[rt] * (len - (len >> 1));    lazy[rt<<1] += lazy[rt];    tree[rt<<1|1] += lazy[rt] * (len >> 1);    lazy[rt<<1|1] += lazy[rt];    lazy[rt] = 0;}void build(int rt = 1, int l = 1, int r = n) {    if (l == r) {        scanf("%lld", &tree[rt]);        return;    }    int m = (l + r) >> 1;    build(lchild);    build(rchild);    push_up(rt);}void update(int L, int R, int delta, int rt = 1, int l = 1, int r = n) {    if (L <= l && r <= R) {        tree[rt] += delta * (r - l + 1);        lazy[rt] += delta;        return;    }    if (lazy[rt]) {        push_down(rt, r - l + 1);    }    int m = (l + r) >> 1;    if (L <= m) {        update(L, R, delta, lchild);    }    if (R > m) {        update(L, R, delta, rchild);    }    push_up(rt);}long long query(int L, int R, int rt = 1, int l = 1, int r = n) {    if (L <= l && r <= R) {        return tree[rt];    }    if (lazy[rt]) {        push_down(rt, r - l + 1);    }    int m = (l + r) >> 1;    long long ret = 0;    if (L <= m) {        ret += query(L, R, lchild);    }    if (R > m) {        ret += query(L, R, rchild);    }    return ret;}int main(int argc, char const *argv[]) {    int q;    scanf("%d%d", &n, &q);    build(1, 1, n);    for (int i = 0; i < q; i++) {        getchar();        int ch;        if ((ch = getchar()) == 'Q') {            int x, y;            scanf("%d%d", &x, &y);            printf("%lld\n", query(x, y));        } else {            int x, y, z;            scanf("%d%d%d", &x, &y, &z);            update(x, y, z);        }    }    return 0;}