[Zhx] 无题
https://www.luogu.org/problemnew/show/T15368
区间修改,区间查询k(<= 10)大值
应该也可以用分块写
#include <cstdio>
#include<algorithm>
#include<iostream>
#include<cstring> #define N 100010
#define tq getchar() using namespace std; struct Note {
long long nm[];
int top;
Note() {memset(nm,,sizeof(nm));top = ;}
} note[N << ],zero_;
long long lazy[N << ];
int n, m; inline int read(){
int x = ; char c = tq;
while(c < '' || c > '') c = tq;
while(c >= '' && c <= '') x = x * + c - '', c = tq;
return x;
} inline Note mer(Note a, Note b) {
if(a.top == ) return b;
if(b.top == ) return a;
Note ans;
int ln = ,rn = ;
while(ln < a.top && rn < b.top) {
if(a.nm[ln] > b.nm[rn]) {ans.nm[ans.top] = a.nm[ln]; ln ++;}
else {ans.nm[ans.top] = b.nm[rn]; rn ++;}
ans.top ++;
if(ans.top == )break;
}
while(ln < a.top) {
if(ans.top == ) break;
ans.nm[ans.top] = a.nm[ln];
ans.top ++;
ln ++;
}
while(rn < b.top) {
if(ans.top == ) break;
ans.nm[ans.top] = b.nm[rn];
ans.top ++;
rn ++;
}
return ans;
} inline void Update(int now) {note[now] = mer(note[now << ],note[now << | ]);} inline void pushdown(int now) {
if(lazy[now]) {
lazy[now << ] += lazy[now];
lazy[now << | ] += lazy[now];
int lson = now << ;
int rson = now << | ;
for(int i = ; i < note[lson].top; i++) note[lson].nm[i] += lazy[now];
for(int i = ; i < note[rson].top; i++) note[rson].nm[i] += lazy[now];
lazy[now] = ;
}
} void build(int l, int r, int now) {
if(l == r) {
note[now].top = ;
note[now].nm[] = read();
return;
}
int mid = (l + r) >> ;
build(l, mid, now << );
build(mid + , r, now << | );
Update(now);
} Note query(int l, int r, int now, int ln, int rn) {
if(l > rn || r < ln) return zero_;
if(l >= ln && r <= rn) return note[now];
pushdown(now);
int mid = (l + r) >> ;
Note lx = query(l, mid, now << , ln, rn);
Note rx = query(mid + , r, now << | , ln, rn);
return mer(lx,rx);
} void add(int l, int r, int now, int ln, int rn, int v) {
if(l > rn || r < ln) return;
if(l >= ln && r <= rn) {
lazy[now] += v;
for(int i = ; i < note[now].top; i++) note[now].nm[i] += v;
return;
}
pushdown(now);
int mid = (l + r) >> ;
add(l, mid, now << , ln, rn, v);
add(mid + , r, now << | , ln, rn, v);
Update(now);
} int main()
{
n = read();
m = read();
build(,n,);
while(m--){
int op = read();
if(op == ) {
int l = read(),r = read(),k = read();
Note zz = query(, n, , l, r);
if(zz.top < k) printf("-1\n");
else cout << zz.nm[k - ] <<"\n";
}
if(op == ) {int l = read(), r = read(), v = read(); add(, n, , l, r, v);}
}
return ;
}
[Zhx] 无题的更多相关文章
- zhx and contest (枚举 + dfs)
zhx and contest Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) ...
- zhx's contest (矩阵快速幂 + 数学推论)
zhx's contest Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) To ...
- HDU - 5187 - zhx's contest (高速幂+高速乘)
zhx's contest Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) To ...
- HDOJ 5188 zhx and contest 贪婪+01背包
zhx and contest Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) ...
- HDU - 5186 - zhx's submissions (精密塔尔苏斯)
zhx's submissions Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others ...
- HDU 5186 zhx's submissions (进制转换)
Problem Description As one of the most powerful brushes, zhx submits a lot of code on many oj and mo ...
- 【HDU5187】zhx's contest
[问题描述] 作为史上最强的刷子之一,zhx的老师让他给学弟(mei)们出n道题.zhx认为第i道题的难度就是i.他想要让这些题目排列起来很漂亮. zhx认为一个漂亮的序列{ai}下列两个条件均需满足 ...
- HDU5187 zhx's contest
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Total Submission ...
- HDU 1871 无题
无题 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submiss ...
随机推荐
- Kubernetes 学习笔记(二):本地部署一个 kubernetes 集群
前言 前面用到过的 minikube 只是一个单节点的 k8s 集群,这对于学习而言是不够的.我们需要有一个多节点集群,才能用到各种调度/监控功能.而且单节点只能是一个加引号的"集群&quo ...
- js复制内容到粘贴板
点击右边内容:<span onclick="copyContent(this);" title="点击复制">啊,我被复制了</span> ...
- POJ 3233-Matrix Power Series( S = A + A^2 + A^3 + … + A^k 矩阵快速幂取模)
Matrix Power Series Time Limit: 3000MS Memory Limit: 131072K Total Submissions: 20309 Accepted: ...
- Effective Java 读书笔记(四):泛型
1 不要使用原始类型 (1)术语 术语 例子 参数化类型(Parameterized type) List<String> 实际类型参数(Actual type parameter) St ...
- jquery.fileupload源码解读笔记
基础编程风格 新建 test.html 和 test.js和 main.js和 无论哪种顺序 <body> <script src="/Sandeep/js/jquery ...
- MP4数据封装格式
一 .MP4 https://blog.csdn.net/sdsszk/article/details/90719075 MP4 由很多个ATOM 嵌套构成,主要的ATOM包括 [ftyp] ...
- H5新增input标签
1.电子邮件 <input type="email" name="email"/> 默认正则:输入内容必须有@符号,@后面必须有内容 2.搜索框 & ...
- CentOS7.X版本系统的下载和安装
一.下载CentOS镜像 1.打开浏览器输入centos.org 2.选择版本下载 3.进入下载页面选择下载版本的种子链接,在迅雷下载即可. 二安装CentOS系统 1.服务器开机,根据界面提示进入磁 ...
- 写Django项目的开发工具或者编辑器
pycharm是写django的最好的编辑器,优先使用pycharm.到官网下载pycharm →https://www.jetbrains.com/pycharm/download→ pycharm ...
- C++——Inheritence
一种错误的观念: 子类继承父类,只把父类的公有成员继承下来,私有的不会继承. 事实上无论是如何继承,都会把父类的所有成员继承下来. #include<iostream> using nam ...