传送门:点击打开链接 题意:一个三元组假设满足j=i+1,k=j+1,ai<=aj<=ak,那么就好的.如今告诉你序列.然后Q次询问.每次询问一个区间[l,r],问区间里有多少个三元组满足要求 思路:刚開始看错题目了,原来三元组是连续3个,这作为bc最后一题也太水了把. . . 先一遍预处理.把连续3个满足条件的找出来,放到还有一个数组里排序去重,用这个数组来给三元组哈希.再扫一遍给三元组在之前那个排序好的数组里二分一下得到下标,大概就是哈希一下,用一个数字来表示. 之后的查询.事实上就是.在…

xiaoxin and his watermelon candy 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5654 Description During his six grade summer vacation, xiaoxin got lots of watermelon candies from his leader when he did his internship at Tencent. Each watermelon cand…

pid=5654">[HDOJ 5654] xiaoxin and his watermelon candy(离线+树状数组) xiaoxin and his watermelon candy Time Limit: 4000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 233    Accepted Submission(s): 61 Problem Des…

xiaoxin and his watermelon candy Problem Description During his six grade summer vacation, xiaoxin got lots of watermelon candies from his leader when he did his internship at Tencent. Each watermelon candy has it's sweetness which denoted by an inte…

前言 听说是线段树离线查询?? 做题做着做着慢慢对离线操作有点感觉了,不过也还没参透,等再做些题目再来讨论离线.在线操作. 这题赛后看代码发现有人用的树状数组,$tql$.当然能用树状数组写的线段树也能写,最重要的还是思维上面. 我当时是怎么想来着.一看异或和前缀和不是很像么,我先把数组中所有数字连续异或并存起来,就像处理前缀和一样的.之后对于每组查询,我只要找到区间中相等的数就可以了,然后再取这些数中相隔最近的距离输出即可.想法很好,但是现实很骨感,这样一来每个区间都这样处理不就超时了么,遂放…

Problem Description During his six grade summer vacation, xiaoxin got lots of watermelon candies from his leader when he did his internship at Tencent. Each watermelon candy has it's sweetness which denoted by an integer number. xiaoxin is very smart…

题意: 给n个编号,m个查询每个查询l,r,求下标区间[l,r]中能分成标号连续的组数(一组内的标号是连续的) 分析: 我们认为初始,每个标号为一个组(线段树维护区间组数),从左向右扫序列,当前标号,要考虑和他相邻的标号的位置,若前面位置出现了和它相邻的标号, 则前面位置组数减一(因为可以合并成一组),查询区间离线处理,保证了查询的正确. #include <map> #include <set> #include <list> #include <cmath&g…

题意: 给你n个数的序列a,q个询问,每个询问给l,r,求在下标i在[l,r]的区间任意两个数的最大公约数中的最大值 分析: 有了hdu3333经验,我们从左向右扫序列,如果当前数的约数在前面出现过,那这个约数可能就是最大的答案.所以我们枚举当前数的所有约数,用线段树维护区间最大值,查询序列离线处理保证查询的正确. #include <map> #include <set> #include <list> #include <cmath> #include…

题意略. 离线处理,离散化.然后就是简单的线段树了.需要根据mod 5的值来维护.具体看代码了. /* 线段树+离散化+离线处理 */ #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> using namespace std; typedef long long ll; #define N 100010 ll sum[N<<2][5]; int a…

K-query Given a sequence of n numbers a1, a2, ..., an and a number of k- queries. A k-query is a triple (i, j, k) (1 ≤ i ≤ j ≤ n). For each k-query (i, j, k), you have to return the number of elements greater than k in the subsequence ai, ai+1, ...,…