DESCRIPTION There are nn buildings lined up, and the height of the ii-th house is hihi. An inteval [l,r][l,r](l≤r)(l≤r) is harmonious if and only if max(hl,…,hr)−min(hl,…,hr)≤kmax(hl,…,hr)−min(hl,…,hr)≤k. Now you need to calculate the number of harmo…

“玲珑杯”ACM比赛 Round #19 Start Time:2017-07-29 14:00:00 End Time:2017-07-29 16:30:00 Refresh Time:2017-07-29 16:42:55 Private B -- Buildings Time Limit:2s Memory Limit:128MByte Submissions:590Solved:151 DESCRIPTION There are nn buildings lined up, and th…

Time Limit:2s Memory Limit:128MByte Submissions:1599Solved:270 DESCRIPTION You have a sequence anan, which satisfies: Now you should find the value of ⌊10an⌋⌊10an⌋. INPUT The input includes multiple test cases. The number of test case is less than 10…

A -- simple math problem Time Limit:2s Memory Limit:128MByte Submissions:1599Solved:270 SAMPLE INPUT 5 20 1314 SAMPLE OUTPUT 5 21 1317 SOLUTION “玲珑杯”ACM比赛 Round #19 题目链接:http://www.ifrog.cc/acm/problem/1145 分析: 这个题解是官方写法,官方代码如下: #include <iostream>…

Interviewe Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6722    Accepted Submission(s): 1592 Problem Description YaoYao has a company and he wants to employ m people recently. Since his compa…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5289 题意:给你n个数和k,求有多少的区间使得区间内部任意两个数的差值小于k,输出符合要求的区间个数 思路:求出区间的最大最小值,只要他们的差值小于k,那么这个区间就符合要求,但是由于n较大,用暴力一定超时,所以就要用别的方法了:而RMQ是可以求区间的最值的,而且预处理的复杂度只有O(nlogn),而查询只是O(1)处理,这样相对来说节约了时间,再根据右端点来二分枚举左端点(其实不用二分好像更快,估…

Interviewe Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 4543    Accepted Submission(s): 1108 Problem Description YaoYao has a company and he wants to employ m people recently. Since his compa…

原文地址:http://www.cnblogs.com/GXZlegend/p/6825389.html 题目描述 小T与小L终于决定走在一起,他们不想浪费在一起的每一分每一秒,所以他们决定每天早上一同晨练来享受在一起的时光. 他们画出了晨练路线的草图,眼尖的小T发现可以用树来描绘这个草图. 他们不愿枯燥的每天从同一个地方开始他们的锻炼,所以他们准备给起点标号后顺序地从每个起点开始(第一天从起点一开始,第二天从起点二开始……). 而且他们给每条道路定上一个幸福的值.很显然他们每次出发都想走幸福值…

Assignment Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 557    Accepted Submission(s): 280 Problem Description Tom owns a company and he is the boss. There are n staffs which are numbered fr…

A -- A simple math problem Time Limit:2s Memory Limit:128MByte Submissions:1599Solved:270 DESCRIPTION You have a sequence anan, which satisfies: Now you should find the value of ⌊10an⌋⌊10an⌋. INPUT The input includes multiple test cases. The number o…

1149 - Buildings Time Limit:2s Memory Limit:128MByte Submissions:588Solved:151 DESCRIPTION There are nn buildings lined up, and the height of the ii-th house is hihi. An inteval [l,r][l,r](l≤r)(l≤r) is harmonious if and only if max(hl,…,hr)−min(hl,…,…

E -- Expected value of the expression DESCRIPTION You are given an expression: A0O1A1O2A2⋯OnAnA0O1A1O2A2⋯OnAn, where Ai(0≤i≤n)Ai(0≤i≤n) represents number, Oi(1≤i≤n)Oi(1≤i≤n) represents operator. There are three operators, &,|,^&,|,^, which means a…

题目链接: B. Strip time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Alexandra has a paper strip with n numbers on it. Let's call them ai from left to right. Now Alexandra wants to split it into…

Friends and Subsequences 题目链接: http://acm.hust.edu.cn/vjudge/contest/121333#problem/H Description Mike and !Mike are old childhood rivals, they are opposite in everything they do, except programming. Today they have a problem they cannot solve on the…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5726 给你n个数,q个询问,每个询问问你有多少对l r的gcd(a[l] , ... , a[r]) 等于的gcd(a[l'] ,..., a[r']). 先用RMQ预处理gcd,dp[i][j] 表示从i开始2^j个数的gcd. 然后用map存取某个gcd所对应的l r的数量. 我们可以在询问前进行预处理,先枚举i,以i为左端点的gcd(a[i],..., a[r])的种类数不会超过log2(n)…

链接:http://acm.hdu.edu.cn/showproblem.php?pid=5726 题意:有N(N <= 100,000),之后有Q(Q <= 100,000)个区间查询[l,r].问ans1 = gcd(al,al+1,...ar) = ?,并且有多少组[l',r'] 的gcd值等于ans1? 思路: 对于求解ans1,由于gcd(a,b,c) = gcd( gcd(a,b),c) 所以可以使用ST表的思想,倍增DP求解区间的gcd值,时间复杂度为O(nlog(n)),之后R…

3473: 字符串 Description 给定n个字符串,询问每个字符串有多少子串(不包括空串)是所有n个字符串中至少k个字符串的子串? Input 第一行两个整数n,k. 接下来n行每行一个字符串. Output 一行n个整数,第i个整数表示第i个字符串的答案. Sample Input 3 1 abc a ab Sample Output 6 1 3 HINT 对于 100% 的数据,1<=n,k<=10^5,所有字符串总长不超过10^5,字符串只包含小写字母. [分析] 这道题用后缀数…

题目大意:给出两个数组,求第一个数组区间内的最大值和第二个区间内的最小值相同的区间有多少种. 题目思路:通过预处理(O(n*Logn))后,每次查询的时间复杂度为O(1),但是如果暴力查询O(n*n)依然会超时,因为最大值,最小值单调性的原因,通过二分来优化查询过程. 首先是预处理,用动态规划(DP)解决.设A[i]是要求区间最值的数列,F[i, j]表示从第i个数起连续2^j个数中的最大值.例如数列3 2 4 5 6 8 1 2 9 7,F[1,0]表示第1个数起,长度为2^0=1的最大值,其…

GCD Time Limit:5000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Submit Status Description Give you a sequence of N(N≤100,000) integers : a1,...,an(0<ai≤1000,000,000). There are Q(Q≤100,000) queries. For each query l,r you have to c…

RMQ存的是区间GCD,然后遍历 i: 1->n, 然后不断地对[i, R]区间进行二分求以i为起点的相同gcd的区间范围,慢慢缩减区间. #include<bits/stdc++.h> #define LL long long using namespace std; ; ], mm[maxn]; map<int, LL>cnt; void init(int n){ mm[] = -; ; i <= n; i ++){ mm[i] = (i&(i - ))?m…

http://acm.split.hdu.edu.cn/showproblem.php?pid=5726 题意:给出一串数字,现在有多次询问,每次询问输出(l,r)范围内所有数的gcd值,并且输出有多少数量区间的gcd值等于该gcd值. 思路: 第一问的话可以用线段树或RMQ来解决,RMQ的话简单点. 有意思的是第二问,假设我们现在固定左端点,那么往右端扩大区间时,gcd单调不增,并且每次至少减少一倍,所以我们可以二分枚举. #include<iostream> #include<alg…

题目链接: Assignment  题意: 给出一个数列,问其中存在多少连续子序列,使得子序列的最大值-最小值<k. 题解: RMQ先处理出每个区间的最大值和最小值(复杂度为:n×logn),相当于求出了每个区间的最大值-最小值.那么现在我们枚举左端点,二分右端点就可以在n×logn×logn的时间内过. #include<bits/stdc++.h> using namespace std; ; int vec[MAX_N]; ]; ]; ; int N,M,T; void ST(in…

[BZOJ4149][AMPPZ2014]Global Warming Description 给定一个序列a[1],a[2],...,a[n].请从中选出一段连续子序列,使得该区间最小值唯一.最大值也唯一. 输出选出的子序列的长度的最大值以及取到最大值时左端点的最小值. Input 第一行包含一个正整数n(1<=n<=500000),表示序列长度. 第二行包含n个正整数,依次表示a[1],a[2],...,a[n](-10^9<=a[i]<=10^9). Output 包含一行两…

Problem Description Give you a sequence of N(N≤100,000) integers : a1,...,an(0<ai≤1000,000,000). There are Q(Q≤100,000) queries. For each query l,r you have to calculate gcd(al,,al+1,...,ar) and count the number of pairs(l′,r′)(1≤l<r≤N)such that gcd…

题目链接: http://poj.org/problem?id=2452 题意:在区间[1,n]上找到满足 a[i]<a[k]<a[j] (i<=k<=j) 的最大子区间 (j-i)如不存在输出 -1. 思路:枚举i,找到 i右边第一个不大于(不是小于) a[i]的数a[k](二分查找+RMQ某段区间的最小值是否小于a[i].最后确定到一个点),于是我们可以得到在区间[i,k-1]范围内的数都会大于 a[i] ,所以对于下标i,它对应的最长区间必定在[i,k-1]之间. 所以,我们…

Tom owns a company and he is the boss. There are n staffs which are numbered from 1 to n in this company, and every staff has a ability. Now, Tom is going to assign a special task to some staffs who were in the same group. In a group, the difference…

不知道错在哪了,求大神指教!!! 思路:用manacher求出每个以str[i]为中心轴的回文串的长度,RMQ预处理区间最大值,对于每个查询,二分最大回文串长,判定是否可行. #include <cstdio> #include <cstring> #include <cstdlib> #include <algorithm> using namespace std; ; char str[ MAXN ]; ]; ]; ][]; int n, len; vo…

https://ac.nowcoder.com/acm/contest/700/I 二维RMQ,贴个板子,注意爆内存,用char就可以了,char也可以存负数. 然后二分枚举对角线长度,理由很简单. 矩阵变大,极值只会变大不会变小.满足单调性. #include<bits/stdc++.h> using namespace std; #define MAXN 501 int n,m,g; int rec[MAXN][MAXN]; char dp[MAXN][MAXN][11][11]; cha…

思路: 对当前值查找最近满足位置: 利用RMQ求出区间最大最小值,再枚举右端点,二分区间找到满足要求的最大区间累加…

因为那时候没怎么补所以就分到了未搞分组里!!!然后因为标题如此之屌吧= =点击量很高,然后写的是无思路,23333,估计看题人真的是觉得博主就是个撒缺.废话不多说了,补题... update////2016/10/3-19:03: 题意: 第一个数求给定询问区间的GCD,第二个数求在这个给定区间里面有多少种连续区间的GCD是等于第一个数. 思路: 区间GCD个数可以利用线段树,或者RMQ搞:鉴于RMQ的查询是O(1),所以RMQ: 求各种GCD的区间数量,预处理掉: (下面这种说是二分,其实还是…