A Simple Problem with Integers（树状数组区间变化和区间求和）

【A Simple Problem with Integers（树状数组区间变化和区间求和）】的更多相关文章

A Simple Problem with Integers(树状数组HDU4267)

A Simple Problem with Integers Time Limit: 5000/1500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4494 Accepted Submission(s): 1384 Problem Description Let A1, A2, - , AN be N elements. You need to deal with two kin…

HDU 4267 A Simple Problem with Integers --树状数组

题意:给一个序列,操作1:给区间[a,b]中(i-a)%k==0的位置 i 的值都加上val 操作2:查询 i 位置的值解法:树状数组记录更新值. 由 (i-a)%k == 0 得知 i%k == a%k,又因为k <= 10,想到建55棵树状数组,即对每个(k,x%k)都建一棵树状数组,每次更新时,在第(k,a%k)棵树状数组上更新a这个点,更新值为val,然后再b+1处更新值为-val,即在[a,b]内更新了val. 查询pos的时候,求出每一个树状数组(k,pos%k)的sum值即可.…

POJ3468 A Simple Problem With Integers 树状数组区间更新区间询问

今天学了很多关于树状数组的技巧.一个是利用树状数组可以简单的实现段更新,点询问(二维的段更新点询问也可以),每次修改只需要修改2个角或者4个角就可以了,另外一个技巧就是这题,原本用线段树做,现在可以用树状数组做的题,只需多维护一个bit即可.具体的思路见下面的链接: http://hi.baidu.com/billdu/item/053f6a15ca301b0a8ebde400 要理解里面的橙色块求的时候是打竖看的,不是打横看的. #pragma warning(disable:4996) #i…

POJ3468 A Simple Problem with Interger [树状数组，差分]

题目传送门 A Simple Problem with Integers Time Limit: 5000MS Memory Limit: 131072K Total Submissions: 130735 Accepted: 40585 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_树状数组

Problem Description Let A1, A2, ... , AN be N elements. You need to deal with two kinds of operations. One type of operation is to add a given number to a few numbers in a given interval. The other is to query the value of some element. Input There…

luogu 2519 [HAOI2011]problem a 动态规划+树状数组

发现每一次 $[b[i]+1,n-a[i]]$ 这个区间的分数必须相同,否则不合法. 而一个相同的区间 $[l,r]$ 最多只能出现区间长度次. 于是,就得到了一个 $dp:$ 将每一种区间的出现次数看作是价值,要选出若干个互不相交的区间使得价值最大. 这个直接用树状数组优化 dp 跑一下就行了~ #include <bits/stdc++.h> #define N 100004 #define setIO(s) freopen(s".in","r",s…

【ZOJ2112】【整体二分+树状数组】带修改区间第k大

The Company Dynamic Rankings has developed a new kind of computer that is no longer satisfied with the query like to simply find the k-th smallest number of the given N numbers. They have developed a more powerful system such that for N numbers a[1],…