Problem 11: Switching Lights [LongFan, 2008]

Farmer John tries to keep the cows sharp by letting them play with
intellectual toys. One of the larger toys is the lights in the barn.
Each of the N (2 <= N <= 500) cow stalls conveniently numbered
1..N has a colorful light above it. At the beginning of the evening, all the lights are off. The cows
control the lights with a set of N pushbutton switches that toggle
the lights; pushing switch i changes the state of light i from off
to on or from on to off. The cows read and execute a list of M (1 <= M <= 2,000) operations
expressed as one of two integers (0 <= operation <= 1). The first operation (denoted by a 0 command) includes two subsequent
integers S_i and E_i (1 <= S_i <= E_i <= N) that indicate a starting
switch and ending switch. They execute the operation by pushing
each pushbutton from S_i through E_i inclusive exactly once. The second operation (denoted by a 1 command) asks the cows to count
how many lights are on in the range given by two integers S_i and
E_i (1 <= S_i <= E_i <= N) which specify the inclusive range in
which the cows should count the number of lights that are on. Help FJ ensure the cows are getting the correct answer by processing
the list and producing the proper counts. PROBLEM NAME: swtch INPUT FORMAT: * Line 1: Two space-separated integers: N and M * Lines 2..M+1: Each line represents an operation with three
space-separated integers: operation, S_i, and E_i SAMPLE INPUT (file swtch.in): 4 5
0 1 2
0 2 4
1 2 3
0 2 4
1 1 4 INPUT DETAILS: Four lights; five commands. Here is the sequence that should
be processed:
Lights
1 2 3 4
Init: O O O O O = off * = on
0 1 2 -> * * O O toggle lights 1 and 2
0 2 4 -> * O * *
1 2 3 -> 1 count the number of lit lights in range 2..3
0 2 4 -> * * O O toggle lights 2, 3, and 4
1 1 4 -> 2 count the number of lit lights in the range 1..4 OUTPUT FORMAT: * Lines 1..number of queries: For each output query, print the count
as an integer by itself on a single line. SAMPLE OUTPUT (file swtch.out): 1
2
就是一开始所有的灯是灭着的,然后.... 可以用线段树解决
/* ***********************************************
Author :guanjun
Created Time :2015/10/4 15:30:49
File Name :1.cpp
************************************************ */
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <stdio.h>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <iomanip>
#include <list>
#include <deque>
#include <stack>
#define ull unsigned long long
#define ll long long
#define mod 90001
#define INF 0x3f3f3f3f
#define maxn 100000+10
#define cle(a) memset(a,0,sizeof(a))
#define ls i<<1
#define rs i<<1|1
const ull inf = 1LL << ;
const double eps=1e-;
using namespace std;
struct node{
int l,r,c;
int sum;
int dist(){
return r-l+;
}
}nod[maxn*];
void push_up(int i){
nod[i].sum=nod[ls].sum+nod[rs].sum;
}
void push_down(int i){
if(nod[i].c){
nod[ls].c^=;
nod[rs].c^=;
nod[ls].sum=nod[ls].dist()-nod[ls].sum;
nod[rs].sum=nod[rs].dist()-nod[rs].sum;
nod[i].c=;
}
}
void build(int i,int l,int r){
nod[i].l=l;
nod[i].r=r;
nod[i].c=nod[i].sum=;
if(l==r){
return ;
}
int mid=(l+r)/;
build(ls,l,mid);
build(rs,mid+,r);
push_up(i);
}
void update(int i,int l,int r){
if(nod[i].l==l&&nod[i].r==r){
nod[i].c^=;
nod[i].sum=nod[i].dist()-nod[i].sum;
return ;
}
push_down(i);
int mid=(nod[i].l+nod[i].r)/;
if(r<=mid)update(ls,l,r);
else if(l>mid)update(rs,l,r);
else {
update(ls,l,mid);
update(rs,mid+,r);
}
push_up(i);
}
int query(int i,int l,int r){
if(nod[i].l==l&&nod[i].r==r){
return nod[i].sum;
}
push_down(i);
int mid=(nod[i].l+nod[i].r)/;
//int sum=0;
if(r<=mid)return query(ls,l,r);
else if(l>mid)return query(rs,l,r);
else return query(ls,l,mid)+query(rs,mid+,r);
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("in.txt","r",stdin);
#endif
//freopen("out.txt","w",stdout);
int n,m,po,x,y;
while(cin>>n>>m){
build(,,n);
for(int j=;j<=m;j++){
scanf("%d%d%d",&po,&x,&y);
if(po==)update(,x,y);
else printf("%d\n",query(,x,y));
}
}
return ;
}

 数据 http://contest.usaco.org/TESTDATA/NOV08_1.htm

usaco2008 nov 区间异或求和的更多相关文章

  1. BZOJ 1230 [Usaco2008 Nov]lites 开关灯:线段树异或

    题目链接:http://www.lydsy.com/JudgeOnline/problem.php?id=1230 题意: 有n盏灯,一开始全是关着的. 有m次操作(p,a,b).p为0,则将区间[a ...

  2. 1230: [Usaco2008 Nov]lites 开关灯

    1230: [Usaco2008 Nov]lites 开关灯 Time Limit: 10 Sec  Memory Limit: 162 MBSubmit: 1162  Solved: 589[Sub ...

  3. NOJ——1669xor的难题(详细的树状数组扩展—异或求和)

    [1669] xor的难题 时间限制: 1000 ms 内存限制: 65535 K 问题描述 最近Alex学长有个问题被困扰了很久,就是有同学给他n个数,然后给你m个查询,然后每个查询给你l和r(左下 ...

  4. bzoj 2819 Nim dfn序+树状数组维护区间异或值

    题目大意 著名游戏设计师vfleaking,最近迷上了Nim.普通的Nim游戏为:两个人进行游戏,N堆石子,每回合可以取其中某一堆的任意多个,可以取完,但不可以不取.谁不能取谁输.这个游戏是有必胜策略 ...

  5. BZOJ 4017 小 Q 的无敌异或 ( 树状数组、区间异或和、区间异或和之和、按位计贡献思想 )

    题目链接 题意 : 中文题 分析 : 首先引入两篇写的很好的题解 题解一.题解二 听说这种和异或相关区间求和的问题都尽量按位考虑 首先第一问.按二进制位计贡献的话.那么对于第 k 位而言 其贡献 = ...

  6. [Usaco2008 Nov]mixup2 混乱的奶牛 简单状压DP

    1231: [Usaco2008 Nov]mixup2 混乱的奶牛 Time Limit: 10 Sec  Memory Limit: 162 MBSubmit: 685  Solved: 383[S ...

  7. [BZOJ1618][Usaco2008 Nov]Buying Hay 购买干草

    [BZOJ1618][Usaco2008 Nov]Buying Hay 购买干草 试题描述 约翰的干草库存已经告罄,他打算为奶牛们采购H(1≤H≤50000)磅干草. 他知道N(1≤N≤100)个干草 ...

  8. BZOJ 1620: [Usaco2008 Nov]Time Management 时间管理( 二分答案 )

    二分一下答案就好了... --------------------------------------------------------------------------------------- ...

  9. BZOJ 1231: [Usaco2008 Nov]mixup2 混乱的奶牛( dp )

    状压dp dp( x , S ) 表示最后一个是 x , 当前选的奶牛集合为 S , 则状态转移方程 : dp( x , S ) =  Σ dp( i , S - { i } )  ( i ∈ S , ...

随机推荐

  1. 洛谷P3327 - [SDOI2015]约数个数和

    Portal Description 共\(T(T\leq5\times10^4)\)组数据.给出\(n,m(n,m\leq5\times10^4)\),求\[\sum_{i=1}^n\sum_{j= ...

  2. 北京集训TEST16——图片加密(fft+kmp)

    题目: Description CJB天天要跟妹子聊天,可是他对微信的加密算法表示担心:“微信这种加密算法,早就过时了,我发明的加密算法早已风靡全球,安全性天下第一!” CJB是这样加密的:设CJB想 ...

  3. POJ2167 Irrelevant Elements

    Time Limit: 5000MS   Memory Limit: 65536KB   64bit IO Format: %lld & %llu Description Young cryp ...

  4. Docker安装运行Redis

    Mac环境 查询镜像: zhoumatoMBP:~ zhou$ docker search redis NAME DESCRIPTION STARS OFFICIAL AUTOMATED redis ...

  5. 马蜂窝ABTest多层分流系统的设计与实现

      什么是 ABTest 产品的改变不是由我们随便「拍脑袋」得出,而是需要由实际的数据驱动,让用户的反馈来指导我们如何更好地改善服务.正如马蜂窝 CEO 陈罡在接受专访时所说:「有些东西是需要 Sen ...

  6. Ajax向前后台传递json和转换

    学生管理系统MVC模式设计心得: jquery .ajax提交data数据格式 jquery 的Ajax方法提交数据,但是是多个参数,具体data的格式如下: data提交的数据类型为:Object  ...

  7. 深入GCD(三): Dispatch Sources

    何为Dispatch Sources简单来说,dispatch source是一个监视某些类型事件的对象.当这些事件发生时,它自动将一个block放入一个dispatch queue的执行例程中.说的 ...

  8. 关于one-hot encoding思考

    Many learning algorithms either learn a single weight per feature, or they use distances between sam ...

  9. 使用nginx转发不了静态文件

    从django runserver的请求来看,都收不到静态文件请求, 查看firefox的web console请求,发现都是403 然后发现nginx不在当前用户组,并且当前用户的项目的读写权限是7 ...

  10. 【软件project】菜鸟俯瞰软件project

    [背景]初次接触软件project,对软件project不是彻底的了解.但学完一遍软件project,我还是有些感触的. 以下我就对我这阶段的软工学习和理解做一下小小的总结,如有不妥之处.欢迎指正. ...