Codeforces Round #250 (Div. 1) D. The Child and Sequence (线段树)
题目链接:http://codeforces.com/problemset/problem/438/D
给你n个数,m个操作,1操作是查询l到r之间的和,2操作是将l到r之间大于等于x的数xor于x,3操作是将下标为k的数变为x。
注意成段更新的时候,遇到一个区间的最大值还小于x的话就停止更新。
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
typedef __int64 LL;
const int MAXN = 1e5 + ;
struct segtree {
int l , r;
LL sum , Min;
}T[MAXN << ];
LL res; void init(int p , int l , int r) {
int mid = (l + r) >> ;
T[p].l = l , T[p].r = r;
if(l == r) {
scanf("%I64d" , &T[p].sum);
T[p].Min = T[p].sum;
return ;
}
init(p << , l , mid);
init((p << )| , mid + , r);
T[p].sum = T[p << ].sum + T[(p << )|].sum;
T[p].Min = (T[p << ].Min > T[(p << )|].Min ? T[p << ].Min : T[(p << )|].Min);
} void updata(int p , int l , int r , LL x) {
int mid = (T[p].l + T[p].r) >> ;
if(T[p].Min < x) {
return ;
}
if(T[p].l == T[p].r) {
T[p].Min %= x;
T[p].sum %= x;
return ;
}
if(r <= mid) {
updata(p << , l , r , x);
}
else if(l > mid) {
updata((p << )| , l , r , x);
}
else {
updata(p << , l , mid , x);
updata((p << )| , mid + , r , x);
}
T[p].sum = T[p << ].sum + T[(p << )|].sum;
T[p].Min = (T[p << ].Min > T[(p << )|].Min ? T[p << ].Min : T[(p << )|].Min);
} void updata2(int p , int index , LL x) {
int mid = (T[p].l + T[p].r) >> ;
if(T[p].l == T[p].r && T[p].l == index) {
T[p].sum = T[p].Min = x;
return ;
}
if(index <= mid) {
updata2(p << , index , x);
}
else {
updata2((p << )| , index , x);
}
T[p].sum = T[p << ].sum + T[(p << )|].sum;
T[p].Min = (T[p << ].Min > T[(p << )|].Min ? T[p << ].Min : T[(p << )|].Min);
} void query(int p , int l , int r) {
int mid = (T[p].l + T[p].r) >> ;
if(l == T[p].l && T[p].r == r) {
res += (LL)T[p].sum;
return ;
}
if(r <= mid) {
query(p << , l , r);
}
else if(l > mid) {
query((p << )| , l , r);
}
else {
query(p << , l , mid);
query((p << )| , mid + , r);
}
} int main()
{
int n , m , l , r , c;
LL x;
scanf("%d %d" , &n , &m);
init( , , n);
while(m--) {
scanf("%d" , &c);
if(c == ) {
scanf("%d %d" , &l , &r);
res = ;
query( , l , r);
printf("%I64d\n" , res);
}
else if(c == ) {
scanf("%d %d %I64d" , &l , &r , &x);
updata( , l , r , x);
//cout << query(1 , 4 , 4) << endl;
}
else {
scanf("%d %I64d" , &l , &x);
updata2( , l , x);
}
}
}
Codeforces Round #250 (Div. 1) D. The Child and Sequence (线段树)的更多相关文章
- Codeforces Round #250 (Div. 1) D. The Child and Sequence 线段树 区间取摸
D. The Child and Sequence Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest ...
- Codeforces Round #250 (Div. 1) D. The Child and Sequence 线段树 区间求和+点修改+区间取模
D. The Child and Sequence At the children's day, the child came to Picks's house, and messed his h ...
- Codeforces Round #250 (Div. 1) D. The Child and Sequence(线段树)
D. The Child and Sequence time limit per test 4 seconds memory limit per test 256 megabytes input st ...
- Codeforces Round #250 (Div. 1) D. The Child and Sequence
D. The Child and Sequence time limit per test 4 seconds memory limit per test 256 megabytes input st ...
- Codeforces Round #271 (Div. 2) F. Ant colony (RMQ or 线段树)
题目链接:http://codeforces.com/contest/474/problem/F 题意简而言之就是问你区间l到r之间有多少个数能整除区间内除了这个数的其他的数,然后区间长度减去数的个数 ...
- Codeforces Round #332 (Div. 2) C. Day at the Beach 线段树
C. Day at the Beach Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/599/p ...
- Codeforces Round #271 (Div. 2) F题 Ant colony(线段树)
题目地址:http://codeforces.com/contest/474/problem/F 由题意可知,最后能够留下来的一定是区间最小gcd. 那就转化成了该区间内与区间最小gcd数相等的个数. ...
- Codeforces Round #225 (Div. 2) E. Propagating tree dfs序+-线段树
题目链接:点击传送 E. Propagating tree time limit per test 2 seconds memory limit per test 256 megabytes inpu ...
- Codeforces Round #343 (Div. 2) D. Babaei and Birthday Cake 线段树维护dp
D. Babaei and Birthday Cake 题目连接: http://www.codeforces.com/contest/629/problem/D Description As you ...
随机推荐
- Jenkins iOS – Git, xcodebuild, TestFlight
Introduction with Jenkins iOS If you are new to continuous integration for mobile platforms then you ...
- iOS开发:插件记录
进入沙盒的插件 https://github.com/TongeJie/ZLGotoSandboxPlugin 图片提示的插件 https://github.com/ksuther/KSImageNa ...
- json化表单数据
/** * josn化表单数据 * @name baidu.form.json * @function * @grammar baidu.form.json(form[, replacer]) * @ ...
- django --------------------- [必要操作]
基本models 命令: python manage.py validate (验证模型有效性, 记得配置 settings.py - INSTALLED_APPS) python manage.py ...
- Java [Leetcode 136]Single Number
题目描述: Given an array of integers, every element appears twice except for one. Find that single one. ...
- Java [Leetcode 169]Majority Element
题目描述: Given an array of size n, find the majority element. The majority element is the element that ...
- Spring学习之AOP
Spring-AOP(Aspect-orented programming) 在业务流程中插入与业务无关的逻辑,这样的逻辑称为Cross-cutting concerns,将Crossing-cutt ...
- jdbc:oracle:thin:@192.168.3.98:1521:orcl(详解)
整理自互联网 一. jdbc:oracle:thin:@192.168.3.98:1521:orcljdbc:表示采用jdbc方式连接数据库oracle:表示连接的是oracle数据库thin:表示连 ...
- HDU5045-Contest(状压dp)
题意: 有n个学生,m道题,给出每个同学解出m个问题的概率,在解题过程中每个学生的解题数的差不大于1,求最大能解出题目数的期望 分析: n很小,知道用状压,但是比赛没做出来(脑子太死了,有一个限制条件 ...
- GIT 分支的理解
乎所有的版本控制系统都以某种形式支持分支. 使用分支意味着你可以把你的工作从开发主线上分离开来,以免影响开发主线. 在很多版本控制系统中,这是一个略微低效的过程——常常需要完全创建一个源代码目录的副本 ...