Problem Description 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 gr…

Assignment Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 3813    Accepted Submission(s): 1771 Problem Description Tom owns a company and he is the boss. There are n staffs which are numbered…

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…

[题目链接]click here~~ [题目大意]: 给出一个数列,问当中存在多少连续子序列,子序列的最大值-最小值<k [思路]:枚举数列左端点.然后二分枚举右端点,用ST算法求区间最值.(或用单调队列的思路) 代码: #include <bits/stdc++.h> using namespace std; const int N=1e5+10; typedef long long LL; #define Max(a,b) a>b?a:b #define Min(a,b) a&…

HDU 5289 - Assignment http://acm.hdu.edu.cn/showproblem.php?pid=5289 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 t…

题目链接:pid=5289">http://acm.hdu.edu.cn/showproblem.php?pid=5289 题面: Assignment Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 672    Accepted Submission(s): 335 Problem Description Tom owns a…

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…

Assignment Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 627    Accepted Submission(s): 318 Problem Description Tom owns a company and he is the boss. There are n staffs which are numbered fr…

题意: 给一个整数序列,多达10万个,问:有多少个区间满足“区间最大元素与最小元素之差不超过k”.k是给定的. 思路: 如果穷举,有O(n*n)复杂度.可以用ST算法先预处理每个区间最大和最小,O(nlogn).再扫一遍整个序列,两个指针L,R用于这样判断:如果A[L,R]这个区间满足要求,则R++,否则统计ans+=R-L,且l++.因为在[L,R]这个区间是满足要求的,那么以L开头的,[L,L].[L,L+1]...[L,R-1]就都是满足要求的,刚好R-L个. #include <bits…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5289 Problem Description 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 tas…

Assignment 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5289 Description 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…

Function Problem Description   The shorter, the simpler. With this problem, you should be convinced of this truth.    You are given an array A of N postive integers, and M queries in the form (l,r). A function F(l,r) (1≤l≤r≤N) is defined as:F(l,r)={A…

题意:求一段长度为n的序列里有多少个子区间内的最大值减最小值小于k. 解法:RMQ+尺取法或单调队列.RMQ可以用st或者线段树,尺取法以前貌似YY出来过……只是不知道是这个东西…… 设两个标记l和r,对于区间[l, r]如果满足题中条件则ans+=(r - l + 1),然后r右移一位,直到不符合条件,将l左移,直到符合条件. 代码: #include<stdio.h> #include<iostream> #include<algorithm> #include&l…

1.区间是一段的,不是断开的哟 2.代码是看着标程写的 3.枚举左端点,二分右端点流程: watermark/2/text/aHR0cDovL2Jsb2cuY3Nkbi5uZXQv/font/5a6L5L2T/fontsize/400/fill/I0JBQkFCMA==/dissolve/70/gravity/Center" alt=""> #include<cstdio> #include<cstring> #include<cmath&…

Assignment Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 902    Accepted Submission(s): 441 Problem Description Tom owns a company and he is the boss. There are n staffs which are numbered fro…

Problem Description 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 gr…

题意:给T足数据,然后每组一个n和k,表示n个数,k表示最大同意的能力差,接下来n个数表示n个人的能力,求能力差在k之内的区间有几个 分析:维护一个区间的最大值和最小值,使得他们的差小于k,于是採用单调队列 普通单调队列做法: #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> using namespace std; const int maxn = 1e6+5; i…

Assignment Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 617    Accepted Submission(s): 314 Problem Description Tom owns a company and he is the boss. There are n staffs which are numbered fr…

题目传送门 /* 题意:问有几个区间最大值-最小值 < k 解法1:枚举左端点,二分右端点,用RMQ(或树状数组)求区间最值,O(nlog(n))复杂度 解法2:用单调队列维护最值,O(n)复杂度,用法 解法3:尺取法,用mutiset维护最值 */ #include <cstdio> #include <algorithm> #include <cstring> #include <cmath> using namespace std; typede…

题目: http://acm.hdu.edu.cn/showproblem.php?pid=5289 Assignment Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 3175    Accepted Submission(s): 1457 Problem Description Tom owns a company and he i…

HDU 2853 Assignment 题目链接 题意:如今有N个部队和M个任务(M>=N),每一个部队完毕每一个任务有一点的效率,效率越高越好.可是部队已经安排了一定的计划,这时须要我们尽量用最小的变动,使得全部部队效率之和最大.求最小变动的数目和变动后和变动前效率之差. 思路:对于怎样保证改变最小,没思路,看了别人题解,恍然大悟,表示想法很机智 试想,假设能让原来那些匹配边,比其它匹配出来总和同样的权值还大,对结果又不影响,那就简单了,这个看似不能做到,事实上是能够做到的 数字最多选出50个…

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…

题目大意:给你n个数,q次询问,每次询问区间[l, r],问a[i]%a[i + 1] % a[i + 2]...%a[j](j <= r)的值 思路:st预处理维护,再二分区间,复杂度n*(logn)*logn //看看会不会爆int!数组会不会少了一维! //取物问题一定要小心先手胜利的条件 #include <bits/stdc++.h> using namespace std; #define LL long long #define ALL(a) a.begin(), a.en…

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…

YaoYao has a company and he wants to employ m people recently. Since his company is so famous, there are n people coming for the interview. However, YaoYao is so busy that he has no time to interview them by himself. So he decides to select exact m i…

Jimmy’s Assignment Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 464    Accepted Submission(s): 224 Problem Description Jimmy is studying Advanced Graph Algorithms at his university. His most…

Description Jimmy is studying Advanced Graph Algorithms at his university. His most recent assignment is to find a maximum matching in a special kind of graph. This graph is undirected, has N vertices and each vertex has degree 3. Furthermore, the gr…

好题~~ 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,枚举后界~~ 又是一种没见过的方法,太弱了/(ㄒoㄒ)/~~ #include <cstdio> #include <cstdlib> #include <cmath> #include <cstring> #include <iostream> #include <queue> #include <algorithm> #…

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)…

Assignment Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 4316    Accepted Submission(s): 1984 Problem Description Tom owns a company and he is the boss. There are n staffs which are numbered f…