A Simple Tree Problem Time Limit: 3000ms Memory Limit: 65536KB This problem will be judged on ZJU. Original ID: 368664-bit integer IO format: %lld      Java class name: Main Prev Submit Status Statistics Discuss Next Type: None   None Graph Theory 2-…

POJ.3468 A Simple Problem with Integers(线段树 区间更新 区间查询) 题意分析 注意一下懒惰标记,数据部分和更新时的数字都要是long long ,别的没什么大坑. 代码总览 #include <cstdio> #include <cstring> #include <algorithm> #define nmax 200000 using namespace std; struct Tree{ int l,r; long lon…

A Simple Problem with Integers Time Limit: 5000MS   Memory Limit: 131072K Total Submissions: 75541   Accepted: 23286 Case Time Limit: 2000MS Description You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type of…

A Simple Problem with Integers Time Limit: 5000MS   Memory Limit: 131072K Total Submissions: 141093   Accepted: 43762 Case Time Limit: 2000MS Description You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type o…

https://vjudge.net/problem/POJ-3468 线段树区间更新(lazy数组)模板题 #include<iostream> #include<cstdio> #include<queue> #include<cstring> #include<algorithm> #include<cmath> #include<map> #define lson l, m, rt<<1 #define…

A Simple Problem with Integers Time Limit: 5000MS   Memory Limit: 131072K Total Submissions: 112228   Accepted: 34905 Case Time Limit: 2000MS Description You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type o…

A Simple Problem with Integers Time Limit: 5000MS   Memory Limit: 131072K Total Submissions: 139191   Accepted: 43086 Case Time Limit: 2000MS Description You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type o…

A Simple Problem with Integers Time Limit: 5000MS   Memory Limit: 131072K Total Submissions: 92632   Accepted: 28818 Case Time Limit: 2000MS Description You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type of…

A Simple Problem with Integers   Description You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type of operation is to add some given number to each number in a given interval. The other is to ask for the sum o…

题目连接: http://poj.org/problem?id=3468 题目大意: 给出n个数,有两种操作: 1:"C a b c",[a,b]中的每一个数都加上c. 2:"Q a b",求[a,b]中每个数相加的和. 解题思路: 线段树更新到每一个节点的话,由于节点数目和查询次数原因会tle,所以在每一个节点内定义一个标志变量表示当前节点的下一层为更新,每次查询时候有需要的话在更新到下一层. #include <cstdio> #include &l…

题目链接:id=3468http://">http://poj.org/problem? id=3468 A Simple Problem with Integers Time Limit: 5000MS   Memory Limit: 131072K Total Submissions: 83959   Accepted: 25989 Case Time Limit: 2000MS Description You have N integers, A1, A2, ... , AN. Yo…

最基本的线段树的区间更新及查询和 用tag(lazy)数组来“延缓”更新,查询或添加操作必须进行pushdown操作,即把tag从p传到lp和rp并清楚tag[p],既然得往lp和rp递归,那么就可以“顺便”往下传 pushdown操作代码 inline void pushdown(int p, int llen, int rlen) { if (tag[p]) { tag[lp] += tag[p], tag[rp] += tag[p]; tree[lp] += tag[p] * llen;…

题目链接:https://www.lydsy.com/JudgeOnline/problem.php?id=3489 题意概述: 给出一个序列,每次询问一个序列区间中仅出现了一次的数字最大是多少,如果没有的话输出0. N<=100000,M<=200000. 分析: 考试的时候YY了一个可持久化KDtree可惜没有打完(一开始想着一维做最后发现自己真是太天真了hahahaha),最后把它改对了. 把两种方法都介绍一下: 持久化KDtree: 首先我们令last[i]表示A[i]在i的左边最靠右…

http://acm.fzu.edu.cn/problem.php?pid=2171 成段增减,区间求和.add累加更新的次数. #include <iostream> #include <cstdio> #include <cmath> #include <vector> #include <cstring> #include <string> #include <algorithm> #include <stri…

题目链接: 传送门 A Simple Problem with Integers Time Limit: 5000MS     Memory Limit: 131072K Description You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type of operation is to add some given number to each number i…

A Simple Problem with Integers Time Limit: 5000MS   Memory Limit: 131072K Total Submissions: 86780   Accepted: 26950 Case Time Limit: 2000MS Description You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type of…

id=3468">点击打开链接题目链接 A Simple Problem with Integers Time Limit: 5000MS   Memory Limit: 131072K Total Submissions: 63565   Accepted: 19546 Case Time Limit: 2000MS Description You have N integers, A1, A2, ... , AN. You need to deal with two kinds of…

A Simple Problem with Integers Time Limit: 5000MS   Memory Limit: 131072K Total Submissions: 67511   Accepted: 20818 Case Time Limit: 2000MS Description You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type of…

#include <iostream> #include <stdio.h> #include <string.h> #define lson rt<<1,L,mid #define rson rt<<1|1,mid+1,R using namespace std; ; int n,q; long long num[maxn]; struct Node{ long long sum,add; bool lazy; }tree[maxn<&l…

#include<iostream> #include<string> #include<algorithm> #include<cstdlib> #include<cstdio> #include<set> #include<map> #include<vector> #include<cstring> #include<stack> #include<cmath> #in…

Description You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type of operation is to add some given number to each number in a given interval. The other is to ask for the sum of numbers in a given interval. In…

You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type of operation is to add some given number to each number in a given interval. The other is to ask for the sum of numbers in a given interval. Input The firs…

You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type of operation is to add some given number to each number in a given interval. The other is to ask for the sum of numbers in a given interval. Input The firs…

Description You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type of operation is to add some given number to each number in a given interval. The other is to ask for the sum of numbers in a given interval. 题意…

Description You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type of operation is to add some given number to each number in a given interval. The other is to ask for the sum of numbers in a given interval. In…

1.给出了一个序列,你需要处理如下两种询问. "C a b c"表示给[a, b]区间中的值全部增加c (-10000 ≤ c ≤ 10000). "Q a b" 询问[a, b]区间中所有值的和. 2.线段树单点更新太费时,所以使用区间更新 3. #include <cstdio> #define L(root) ((root) << 1) #define R(root) (((root) << 1) + 1) ; int nu…

Time Limit: 5000MS   Memory Limit: 131072K Total Submissions: 115624   Accepted: 35897 Case Time Limit: 2000MS Description You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type of operation is to add some give…

题目地址:POJ 3468 打了个篮球回来果然神经有点冲动. . 无脑的狂交了8次WA..竟然是更新的时候把r-l写成了l-r... 这题就是区间更新裸题. 区间更新就是加一个lazy标记,延迟标记,仅仅有向下查询的时候才将lazy标记向下更新.其它的均按线段树的来即可. 代码例如以下: #include <iostream> #include <cstdio> #include <cstring> #include <math.h> #include &l…

题意:区间add,区间求和. #include<cstdio> #include<cstring> #include<cstdlib> #include<cctype> #include<cmath> #include<iostream> #include<sstream> #include<iterator> #include<algorithm> #include<string> #…

Description Given a rooted tree, each node has a boolean (0 or 1) labeled on it. Initially, all the labels are 0. We define this kind of operation: given a subtree, negate all its labels. And we want to query the numbers of 1's of a subtree. Input Mu…