Codeforces Round #263 (Div. 1) C. Appleman and a Sheet of Paper 树状数组暴力更新
C. Appleman and a Sheet of Paper
Appleman has a very big sheet of paper. This sheet has a form of rectangle with dimensions 1 × n. Your task is help Appleman with folding of such a sheet. Actually, you need to perform q queries. Each query will have one of the following types:
- Fold the sheet of paper at position pi. After this query the leftmost part of the paper with dimensions 1 × pi must be above the rightmost part of the paper with dimensions 1 × ([current width of sheet] - pi).
- Count what is the total width of the paper pieces, if we will make two described later cuts and consider only the pieces between the cuts. We will make one cut at distance li from the left border of the current sheet of paper and the other at distance ri from the left border of the current sheet of paper.
Please look at the explanation of the first test example for better understanding of the problem.
Input
The first line contains two integers: n and q (1 ≤ n ≤ 105; 1 ≤ q ≤ 105) — the width of the paper and the number of queries.
Each of the following q lines contains one of the described queries in the following format:
- "1 pi" (1 ≤ pi < [current width of sheet]) — the first type query.
- "2 li ri" (0 ≤ li < ri ≤ [current width of sheet]) — the second type query.
Output
For each query of the second type, output the answer.
input
7 4
1 3
1 2
2 0 1
2 1 2
output
4
3
思路: 暴力更新, 然后用FenwickTree 或者SegmentTree进行区间求和即可。
因为每个位置上的value只会更新到别的位置一次,所以暴力的话复杂度也是O(n), 然后更新的时候分两种情况, 如果折过去的长度大于右边界 就相当于把右面的对应长度折过来, 否则就是题目中所说的从左面折了。
我用ua, ub,维护了当前区间的左右端点, 每次查询也分两种情况。
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + ;
namespace FenwickTree {
int arr[maxn];
void Modify(int x, int d) {
while(x < maxn) {
arr[x] += d;
x += x & -x;
}
}
void init(int n) {
memset(arr, , sizeof arr);
for(int i = ; i <= n; i++) {
Modify(i, );
}
}
int query(int x) {
int res = ;
while (x > ) {
res += arr[x];
x -= x & -x;
}
return res;
}
}
int main() {
#ifndef ONLINE_JUDGE
freopen("in.txt","r",stdin);
#endif
int n, q;
while (~ scanf ("%d%d", &n, &q)) {
int tot = , direction = ;
FenwickTree::init(n);
int ub = n;
for (int j = ; j < q; j++) {
int op, l, r, p;
int ua = tot+;
scanf ("%d", &op);
if (op == ) {
scanf ("%d", &p);
if (!direction) {
if (*p <= + ub - ua) {
for (int i = ua; i <= ua+p-; i++) {
FenwickTree::Modify(*ua+*p-i-, FenwickTree::query(i) - FenwickTree::query(i-));
}
tot += p;
} else {
p = ub - (ua + p-);
for (int i = ub; i >= ub-p+; i--) {
FenwickTree::Modify(*ub-*p+-i, FenwickTree::query(i) - FenwickTree::query(i-));
}
ub = ub - p;
direction ^= ;
}
}else{
if (*p > + ub - ua){
p = ub - (ua + p-);
for (int i = ua; i <= ua+p-; i++) {
FenwickTree::Modify(*ua+*p-i-, FenwickTree::query(i) - FenwickTree::query(i-));
}
direction ^= ;
tot += p;
}else{
for (int i = ub; i >= ub-p+; i--) {
FenwickTree::Modify(*ub-*p+-i, FenwickTree::query(i) - FenwickTree::query(i-));
}
ub = ub - p;
}
} } else {
scanf ("%d%d", &l, &r);
if (!direction) {
printf("%d\n", FenwickTree::query(ua+r-)-FenwickTree::query(ua-+l));
}else{
printf("%d\n", FenwickTree::query(ub-l)-FenwickTree::query(ub-r));
}
}
}
}
return ;
}
Codeforces Round #263 (Div. 1) C. Appleman and a Sheet of Paper 树状数组暴力更新的更多相关文章
- Codeforces Round #365 (Div. 2) D - Mishka and Interesting sum(离线树状数组)
http://codeforces.com/contest/703/problem/D 题意: 给出一行数,有m次查询,每次查询输出区间内出现次数为偶数次的数字的异或和. 思路: 这儿利用一下异或和的 ...
- Codeforces Round #227 (Div. 2) E. George and Cards set内二分+树状数组
E. George and Cards George is a cat, so he loves playing very much. Vitaly put n cards in a row in ...
- Codeforces Round #261 (Div. 2) D. Pashmak and Parmida's problem (树状数组求逆序数 变形)
题目链接 题意:给出数组A,定义f(l,r,x)为A[]的下标l到r之间,等于x的元素数.i和j符合f(1,i,a[i])>f(j,n,a[j]),求i和j的种类数. 我们可以用map预处理出 ...
- Codeforces Round #381 (Div. 2) D. Alyona and a tree dfs序+树状数组
D. Alyona and a tree time limit per test 2 seconds memory limit per test 256 megabytes input standar ...
- Codeforces Round #590 (Div. 3)【D题:维护26棵树状数组【好题】】
A题 题意:给你 n 个数 , 你需要改变这些数使得这 n 个数的值相等 , 并且要求改变后所有数的和需大于等于原来的所有数字的和 , 然后输出满足题意且改变后最小的数值. AC代码: #includ ...
- 贪心 Codeforces Round #263 (Div. 2) C. Appleman and Toastman
题目传送门 /* 贪心:每次把一个丢掉,选择最小的.累加求和,重复n-1次 */ /************************************************ Author :R ...
- Codeforces Round #348 (VK Cup 2016 Round 2, Div. 2 Edition) E. Little Artem and Time Machine 树状数组
E. Little Artem and Time Machine 题目连接: http://www.codeforces.com/contest/669/problem/E Description L ...
- Codeforces Round #263 (Div. 2) D. Appleman and Tree(树形DP)
题目链接 D. Appleman and Tree time limit per test :2 seconds memory limit per test: 256 megabytes input ...
- Codeforces Round #263 (Div. 2) A. Appleman and Easy Task【地图型搜索/判断一个点四周‘o’的个数的奇偶】
A. Appleman and Easy Task time limit per test 1 second memory limit per test 256 megabytes input sta ...
随机推荐
- Android OpenGL ES 3.0 纹理应用
本文主要演示OpenGL ES 3.0 纹理演示.接口大部分和2.0没什么区别,脚本稍微有了点变化而已. 扩展GLSurfaceView package com.example.gles300; im ...
- C# 如何创建接口以及使用接口的简单Demo(转载!)
//No:1 首先,我们要封装一个接口,接口中不要实现具体的方法(说白了这就是一个架子而已!) using System;using System.Collections.Generic;using ...
- StudioStyle 使用 厌倦了默认的Visutal Studio样式了,到这里找一个酷的试试
厌倦了默认的Visutal Studio样式了,到这里找一个酷的试试 http://studiostyl.es/ 去下载个自己喜欢的编码样式吧 如果你有想法 有能力 可以自己去做一个自己喜欢的 OK ...
- ASP。net中如何在一个按钮click事件中调用另一个按钮的click事件
方法一: 直接指定 事件<asp:Button ID="btn1" runat="server" Text="按钮1" onclick ...
- 后台线程,优先级,sleep,yield
1.后台线程,是指在程序运行的时候在后台提供一种通用服务的线程,并且这种线程并不属于程序中不可获取的部分.当所有非后台线程结束时,程序也就 终止了,同时会杀死进程中所有后台线程.main()是一个非后 ...
- objective-c相关知识点
1,objective-c中实现线程同步: Mutexlock (互斥锁).NSCondition lock (条件锁)消息传送 2,UDP和TCP: TCP :传输控制协议,可以提供面向连接的.可靠 ...
- jQuery 效果- 动画
jQuery animate() 方法允许您创建自定义的动画. jQuery 动画实例 jQuery jQuery 动画 - animate() 方法 jQuery animate() 方法用于创建自 ...
- java问题整理
1.一个“.java”源文件中是否可以包括多个类(不是内部类)?有什么限制? 答:可以有多个类.但只能有一个public类.并且public类名必须与文件名相一致. 2.Java有没有goto? ...
- Codeforces 527E Data Center Drama(欧拉回路)
题意: 给定一个无向图连通图,把这个的无向边变成有向边,并添加最少的有向边使这个图每个结点的出度为偶数. Solution: 题目很长,并且很多条件说的不太直接,确实不太好懂. 首先先看得到的无向图, ...
- Linux 系统命令及其使用详解(大全)
(来源: 中国系统分析员) cat cd chmod chown cp cut 1.名称:cat 使用权限:所有使用者 使用方式:cat [-AbeEnstTuv] [--help] [--versi ...