poj3468 A Simple Problem with Integers 题意:O(-1) 思路:O(-1) 线段树功能:update:成段增减 query:区间求和 Sample Input 10 5 1 2 3 4 5 6 7 8 9 10 Q 4 4 Q 1 10 Q 2 4 C 3 6 3 Q 2 4 Sample Output 4 55 9 15 Hint The sums may exceed the range of 32-bit integers. 题目大意:   N个数 M…

题目传送门 A Simple Problem with Integers Time Limit: 5000MS   Memory Limit: 131072K Total Submissions: 142198   Accepted: 44136 Case Time Limit: 2000MS Description You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One…

A Simple Problem with Integers Time Limit: 5000MS   Memory Limit: 131072K Total Submissions: 92127   Accepted: 28671 Case Time Limit: 2000MS 描述 You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type of operatio…

A Simple Problem with Integers Time Limit: 5000MS   Memory Limit: 131072K Total Submissions: 60745   Accepted: 18522 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: 57666   Accepted: 17546 Case Time Limit: 2000MS Description You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type of…

http://poj.org/problem?id=3468 实现一个线段树,能够做到区间修改和区间查询和. 明显板子题. #include<cstdio> #include<cmath> #include<algorithm> #include<cstring> #include<iostream> using namespace std; typedef long long ll; inline ll read(){ ll X=,w=; ;…

题目链接:https://vjudge.net/problem/POJ-3468 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 nu…

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…

题目链接:http://poj.org/problem?id=3468 这题是线段树的题,拿来学习treap. 不旋转的treap. #include <cstdio> #include <cstring> #include <algorithm> #include <vector> #include <map> #include <set> #include <string> #include <bitset>…

题目地址http://poj.org/problem?id=3468 题目大意很简单,有两个操作,一个 Q a, b 查询区间[a, b]的和 C a, b, c让区间[a, b] 的每一个数+c 第一次线段树的延时标记,花了好大的功夫才写好==! 很容易看出来使用使用线段树记录区间的和,但是难点在于每次修改的是一个区间而不是一个点 所以采用的方法就是每次做修改操作时,只将区间［a,b］的标记+c,而不是真正意义上的将区间[a, b] 的每一个值+c. 而当我们做查询操作时,就只需要将区间[a,…