萌新昧子刚学OI，线段树求调QAQ

kaceqwq @ 2022-08-17 08:05:13

by kaceqwq @ 2022-08-17 08:21:14

@JackMerryYoung 大佬，题目只要求最大值，不用求和，所以 sum 就只是求最大值的。

by fajerwerki @ 2022-08-17 08:23:35

@JackMerryYoung 注意看

他写的是：

tree[p].sum += x;

并非

tree[p].sum += (l - r + 1) * x;

所以只是求最大值

by JackMerryYoung @ 2022-08-17 08:24:04

@neach__joup 太生草了，代码。

by kaceqwq @ 2022-08-17 08:24:47

对不起，我是蒟蒻

by ioker @ 2022-08-17 08:26:20

@kaceqwq 《昧子》

by fajerwerki @ 2022-08-17 08:28:01

@dpkajj 膜拜大佬

by fajerwerki @ 2022-08-17 08:28:43

@dpkajj c

验证码不给你面子

by Micnation_AFO @ 2022-08-17 08:28:50

@dpkajj 改了一下样例过了;

1. update2 里面递归调用的函数你写成 update1 了（

2. push_down 改成了这样：

void Push_down (int p) {
    if (tree[p].flag) {
        tree[p * 2].tag2 = 0;
        tree[p * 2 + 1].tag2 = 0;
        tree[p * 2].tag1 = tree[p].tag1;
        tree[p * 2 + 1].tag1 = tree[p].tag1;
        tree[p * 2].sum = tree[p].tag1;
        tree[p * 2 + 1].sum = tree[p].tag1;
        tree[p * 2].flag = tree[p * 2 + 1].flag = 1;
    } 
    if (tree[p].tag2) {
        tree[p * 2].sum += tree[p].tag2;
        tree[p * 2 + 1].sum += tree[p].tag2;
        tree[p * 2].tag2 += tree[p].tag2;
        tree[p * 2 + 1].tag2 += tree[p].tag2;
    }
    tree[p].flag = tree[p].tag1 = tree[p].tag2 = 0;
}

by ioker @ 2022-08-17 08:30:04

@Leap_hash_jperm

You need to @kaceqwq

by ioker @ 2022-08-17 08:30:19

@neach__joup cd