思路: 对于一个二进制100011: 尽量将填满:填成111111: 然后有一个很好算的方法 gets(n)表示二进制下N有多少位,N^X=(111111)2 X=111111^N; 其实答案可以直接 ANS=N*(N+1);找的规律: #include <cstdlib> #include <cctype> #include <cstring> #include <cstdio> #include <cmath> #include <al…

HDU 3397 Sequence operation 题目链接 题意:给定一个01序列,有5种操作 0 a b [a.b]区间置为0 1 a b [a,b]区间置为1 2 a b [a,b]区间0变成1,1变成0 3 a b 查询[a,b]区间1的个数 4 a b 查询[a,b]区间连续1最长的长度 思路:线段树线段合并.须要两个延迟标记一个置为01,一个翻转,然后因为4操作,须要记录左边最长0.1.右边最长0.1,区间最长0.1,然后区间合并去搞就可以 代码: #include <cstdi…

Sequence II Time Limit: 9000/4500 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 1422    Accepted Submission(s): 362 Problem Description Mr. Frog has an integer sequence of length n, which can be denoted as a1,a2…

题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=5146 Sequence Description Today we have a number sequence A includes n elements.Nero thinks a number sequence A is good only if the sum of its elements with odd index equals to the sum of its elements wi…

任意门:http://acm.hdu.edu.cn/showproblem.php?pid=6395 Sequence Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 2564    Accepted Submission(s): 999 Problem Description Let us define a sequence as…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5312 Sequence Time Limit: 2000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others) Total Submission(s): 1336    Accepted Submission(s): 410 Problem Description Today, Soda has learned a…

Number Sequence Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 15149    Accepted Submission(s): 6644 Problem Description Given two sequences of numbers : a[1], a[2], ...... , a[N], and b[1], b…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3397 题意:给你一个长度为n的0,1序列,支持下列五种操作, 操作0(0 a b):将a到b这个区间的元素全部置为0. 操作1(1 a b):将a到b这个区间的元素全部置为1. 操作2(2 a b):将a到b这个区间所有的0置为1,所有的1置为0. 操作3(3 a b):查询a到b这个区间1的总数. 操作4(4 a b):查询a到b这个区间连续1的最长长度 本题属于简单的区间更新线段树 重点:0操作…

题目链接 http://poj.org/problem?id=1141 Description Let us define a regular brackets sequence in the following way: 1. Empty sequence is a regular sequence. 2. If S is a regular sequence, then (S) and [S] are both regular sequences. 3. If A and B are reg…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1711 数字KMP,原来还能这么用 #include<stdio.h> ],b[]; ]; int n,m; void getNext() { int j,k; j=; k=-; next[]=-; while(j<m) { ||b[j]==b[k]) next[++j]=++k; else k=next[k]; } } //返回首次出现的位置 int KMP_Index() { ,j=; g…