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Four Operations Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 22    Accepted Submission(s): 12 Problem Description Little Ruins is a studious boy, recently he learned the four operations!Now h…

Problem Description Little Ruins is a studious boy, recently he learned the four operations! Now he want to use four operations to generate a number, he takes a string which only contains digits ‘1’ - ‘9’, and split it into 5 intervals and add the fo…

题意:把'+', '-', '*' 和'/'按顺序插入任意两数字间隔,使得操作得到后计算后最大. 思路:没想到是个水题,打的时候想得太复杂了.这道题其实只要考虑*和/.显然我们要把a*b/c弄到最小.那么ab只有一位,c可能有两位.m+n-a*b/c最大,那么mn和最大,那么就是-号前面分割最大和.那么就是max(第一位+后面,前面+最后一位). 代码: #include<set> #include<map> #include<stack> #include<cm…

题目: http://acm.hdu.edu.cn/showproblem.php?pid=6315 Naive Operations Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 502768/502768 K (Java/Others) Problem Description In a galaxy far, far away, there are two integer sequence a and b of length…

Problem DescriptionIn a galaxy far, far away, there are two integer sequence a and b of length n.b is a static permutation of 1 to n. Initially a is filled with zeroes.There are two kind of operations:1. add l r: add one for al,al+1...ar2. query l r:…

Naive Operations Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 502768/502768 K (Java/Others)Total Submission(s): 0    Accepted Submission(s): 0 Problem Description In a galaxy far, far away, there are two integer sequence a and b of length…

Naive Operations Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 502768/502768 K (Java/Others)Total Submission(s): 1791    Accepted Submission(s): 772 Problem Description In a galaxy far, far away, there are two integer sequence a and b of le…

hdu 6315 题意:对于一个数列a,初始为0,每个a[ i ]对应一个b[i],只有在这个数字上加了b[i]次后,a[i]才会+1. 有q次操作,一种是个区间加1,一种是查询a的区间和. 思路:线段树,一开始没用lazy,TLE了,然后开始想lazy的标记,这棵线段树的特点是,每个节点维护 :一个区间某个a 要增加1所需个数的最小值,一个区间已加上的mx的最大值标记,还有就是区间和sum. (自己一开始没有想到mx标记,一度想把lazy传回去. (思路差一点就多开节点标记. #include…

Naive Operations Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 502768/502768 K (Java/Others)Total Submission(s): 2114    Accepted Submission(s): 915 Problem Description In a galaxy far, far away, there are two integer sequence a and b of le…

题意:a数组初始全为0,b数组题目给你,有两种操作: 思路:dls的思路很妙啊,我们可以将a初始化为b,加一操作改为减一,然后我们维护一个最小值,一旦最小值为0,说明至少有一个ai > bi,那么找出所有为0的给他的最终结果加上一并且重置为bi,维护一个区间和,询问时线段树求和.一开始updateMin没加判断,单个复杂度飙到nlog(n),疯狂TLE... 代码: #include<cstdio> #include<vector> #include<stack>…

题意:数量为N的序列a和b,a初始全为0,b为给定的1-N的排列.有两种操作:1.将a序列区间[L,R]中的数全部+1:2.查询区间[L,R]中的 ∑⌊ai/bi⌋(向下取整) 分析:对于一个位置i,如果ai<bi,那么该位置不能对结果做出贡献:而当某一次操作后,ai>=bi了,就对结果的贡献值+1.那么可以用在线段树的结点中维护每个区间的最大a和最小b,和已经产生的贡献cnt.如果一个区间中最大的a超过了b,那么就说明此次更新操作使该区间的结果产生了变化,那么就要向下找到那个产生贡献的位置.…

题意: 给定S,N,把S+1,S+2,...S+N这N个数填到1,2,...,N里,要求X只能填到X的因子的位置.(即X%Y=0,那么X才能放在Y位置) 问是否能够放满. 分析:经过小队的分析得出的结论是如果S+1,S+2,...S+N有两个素数就肯定是不行的对吧 ,虽然素数可以取本身当如果素数可以取到本身s=0||s=1 , 前面可以特判出来 , 所以可以估算一下素数的最大间隔(我取504),N超过必然无解.N小于504的情况下,直接暴力建边(能整除就连边),然后跑二分图匹配即可. 网上有片博…

发现每次区间加只能加1,最多全局加\(n\)次,这样的话,最后的答案是调和级数为\(nlogn\),我们每当答案加1的时候就单点加,最多加\(nlogn\)次,复杂度可以得当保证. 然后问题就是怎么判断答案是否该加1.我们可以用线段树设初值为给出的排列,把区间加改成区间减,维护最小值.当最小值为0是遍历左右子树,找到该加1的节点,一共会找\(nlongn\)次复杂度也可以得到保证. #include<iostream> #include<cstring> #include<c…

<题目链接> 题目大意: 给出两个序列,a序列全部初始化为0,b序列为输入值.然后有两种操作,add x y就是把a数组[x,y]区间内全部+1,query x y是查询[x,y]区间内∑[ai/bi].([ai/bi]代表ai/bi后向下取整) 解题分析: 首先,如果每次+1都暴力更新到每个叶子节点肯定会超时,但是如果不更新到叶子节点又不好维护每个节点对应区间 ∑[ai/bi] 的值,所以我们可以每个节点都维护四个值.sum值代表这个区间每个节点整数部分的所有数之和,lazy进行懒惰标记,避…

2018 Multi-University Training Contest 2) HDU 6311 Cover HDU 6312 Game HDU 6313 Hack It HDU 6314 Matrix HDU 6315 Naive Operations HDU 6318 Swaps and Inversions…

http://acm.hdu.edu.cn/showproblem.php?pid=6315 题意 a数组初始全为0,b数组为1-n的一个排列.q次操作,一种操作add给a[l...r]加1,另一种操作query查询Σfloor(ai/bi)(i=l...r). 分析 真的是太naive啦,现场时没做出来. 看见区间自然想起线段树,那么这里的关键就是整除问题,只有达到一定数量才会对区间和产生影响. 反过来想,先把a[i]置为b[i],那么每次add时就是-1操作,当a[i]为0时区间和+1,再把…

题目大意 题目链接Naive Operations 题目大意: 区间加1(在a数组中) 区间求ai/bi的和 ai初值全部为0,bi给出,且为n的排列,多组数据(<=5),n,q<=1e5 axmorz 思路 因为是整除,所以一次加法可以对ans 没有影响 当ai是bi的倍数,对ans会有贡献 所以我们维护一个sum,初值为bi(只对于线段树的叶子节点有用) 当区间+1的时候 我们对sum-1 当sum=0的时候(倍数) ans++,sum=bi 然后再维护一个区间最小值 当区间内的mi>…

Problem Description In a galaxy far, far away, there are two integer sequence a and b of length n.b is a static permutation of 1 to n. Initially a is filled with zeroes.There are two kind of operations:1. add l r: add one for al,al+1...ar 2. query l…

Naive Operations Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 502768/502768 K (Java/Others)Total Submission(s): 2691    Accepted Submission(s): 1183 Problem Description In a galaxy far, far away, there are two integer sequence a and b of l…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3584 Cube Problem Description Given an N*N*N cube A, whose elements are either 0 or 1. A[i, j, k] means the number in the i-th row , j-th column and k-th layer. Initially we have A[i, j, k] = 0 (1 <= i, …

题目链接 http://acm.hdu.edu.cn/showproblem.php?pid=5475 Problem Description One day, a useless calculator was being built by Kuros. Let's assume that number X is showed on the screen of calculator. At first, X = 1. This calculator only supports two types…

http://acm.hdu.edu.cn/showproblem.php?pid=4960 2014 Multi-University Training Contest 9 Another OCD Patient Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 181    Accepted Submission(s): 58 Pr…

http://acm.hdu.edu.cn/showproblem.php?pid=1890 Robotic Sort Problem Description Somewhere deep in the Czech Technical University buildings, there are laboratories for examining mechanical and electrical properties of various materials. In one of yest…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4897 Problem Description There is an old country and the king fell in love with a devil. The devil always asks the king to do some crazy things. Although the king used to be wise and beloved by his peopl…

http://acm.hdu.edu.cn/showproblem.php?pid=5818 Joint Stacks Problem Description   A stack is a data structure in which all insertions and deletions of entries are made at one end, called the "top" of the stack. The last entry which is inserted i…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5938 题意:给出一个长度最大是2020的数字串, 你要把数字串划分成55段, 依次填入'+', '-', '*', '/', 问能得到的最大结果. 想让这个结果最大,那尽可能让减号左边的数最大,只需要关心左边的第一位是一个数还是最后一位是一个数,后面的枚举*和/的位置就行了. #include <bits/stdc++.h> using namespace std; typedef long lo…

HDU 5818 Joint Stacks(联合栈) Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Description 题目描述 A stack is a data structure in which all insertions and deletions of entries are made at one end, called the "top" of…

题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=5349 MZL's simple problem Description A simple problemProblem DescriptionYou have a multiple set,and now there are three kinds of operations:1 x : add number x to set2 : delete the minimum number (if the…

题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=3074 Minimum Inversion Number Description Tired of playing computer games, alpc23 is planning to play a game on numbers. Because plus and subtraction is too easy for this gay, he wants to do some multipl…