A Simple Problem with Integers（线段树区间更新模板）

 

最基本的线段树的区间更新及查询和

用tag(lazy)数组来“延缓”更新，查询或添加操作必须进行pushdown操作，即把tag从p传到lp和rp并清楚tag[p]，既然得往lp和rp递归，那么就可以“顺便”往下传

pushdown操作代码

inline void pushdown(int p, int llen, int rlen) {
    if (tag[p]) {
        tag[lp] += tag[p], tag[rp] += tag[p];
        tree[lp] += tag[p] * llen;
        tree[rp] += tag[p] * rlen;
        tag[p] = 0;
    }
} 

还要注意必须用long long来存值

代码如下

#include <cstdio>
#include <algorithm>
#define ll long long
#define lp p<<1
#define rp p<<1|1
using namespace std;

const int maxn = 100000;
ll tree[maxn<<2], tag[maxn<<2];
void build(int p, int l, int r) {
    if (l == r) {
        scanf("%lld", &tree[p]);
        return;
    }
    int mid = (l + r) >> 1;
    build(lp, l, mid); build(rp, mid + 1, r);
    tree[p] = tree[lp] + tree[rp];
}
inline void pushdown(int p, int llen, int rlen) {
    if (tag[p]) {
        tag[lp] += tag[p], tag[rp] += tag[p];
        tree[lp] += tag[p] * llen;
        tree[rp] += tag[p] * rlen;
        tag[p] = 0;
    }
} 
void add(int p, int l, int r, int x, int y, int z) {
    if (x <= l && y >= r) {
        tag[p] += 1LL * z;
        tree[p] += 1LL * z * (r - l + 1);
        return;
    }
    int mid = (l + r) >> 1;
    pushdown(p, mid - l + 1, r - mid);
    if (x <= mid) add(lp, l, mid, x, y, z);
    if (y > mid) add(rp, mid + 1, r, x, y, z);
    tree[p] = tree[lp] + tree[rp];
}

ll find(int p, int l, int r, int x, int y) {
    if (x <= l && y >= r) return tree[p];
    int mid = (l + r) >> 1;
    pushdown(p, mid - l + 1, r - mid);
    if (y <= mid) return find(lp, l, mid, x, y);
    if (x > mid) return find(rp, mid + 1, r, x, y);
    return find(lp, l, mid, x, y) + find(rp, mid + 1, r, x, y);
}

int main() {
    int n, q;
    scanf("%d%d", &n, &q);
    build(1, 1, n);
    while (q--) {
        char s[5];
        int x, y;
        scanf("%s%d%d", s, &x, &y);
        if (s[0] == 'Q') {
            printf("%lld\n", find(1, 1, n, x, y));
        } else {
            int z; scanf("%d", &z);
            add(1, 1, n, x, y, z);
        }
    }
    return 0;
}