题意:查询的是区间内每个数出现次数的平方×该数值的和. 分析:虽然是道莫队裸体,但是姿势不对就会超时.答案可能爆int,所以要开long long 存答案.一开始的维护操作,我先在res里减掉了a[pos]*cnt[a[pos]]*cnt[a[pos]],将cnt[a[pos]]+1,再将乘积加回.其实根据数学原理,K^2和(K+1)^2差值是2K+1,那么其实每次更新时只要加上或减去a[pos]*(2*cnt[pos]+1)即可,这样更高效. #include<bits/stdc++.h>

D. Powerful array time limit per test 5 seconds memory limit per test 256 megabytes input standard input output standard output An array of positive integers a1, a2, ..., an is given. Let us consider its arbitrary subarray al, al + 1..., ar, where 1 

莫队算法就是优化的暴力算法.莫队算法是要把询问先按左端点属于的块排序,再按右端点排序.只是预先知道了所有的询问.可以合理的组织计算每个询问的顺序以此来降低复杂度. D. Powerful array 典型的莫队算法题 #include <iostream> #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <string&g

codefroce D. Powerful array[初识块状数组] 由于是初始所以,仅仅能先用别人的分析.囧... 题目: 给定一个数列:A1, A2,--,An,定义Ks为区间(l,r)中s出现的次数. t个查询,每一个查询l,r,对区间内全部a[i],求sigma(K^2*a[i]) 离线+分块 将n个数分成sqrt(n)块. 对全部询问进行排序,排序标准: 1. Q[i].left /block_size < Q[j].left / block_size (块号优先排序) 2. 假设1

D. Powerful array 题意 给定一个数列:a[i] (1<= i <= n) K[j]表示 在区间 [l,r]中j出现的次数.有t个查询,每个查询l,r,对区间内所有a[i],求sigma(K[a[i]] * K[a[i]] * a[i]). 分析 首先将区间分块,每块sqrt(n)个,先读入所有的查询,对查询进行排序:先按块序号从小到大,再按右端点的值从小大到排序. 按顺序读入查询,动态变更区间,以及区间的值.如果某个点出现了 x 次,出现 x + 1 次时,区间内增加(2 *

题目链接:点击传送 D. Powerful array time limit per test 5 seconds memory limit per test 256 megabytes input standard input output standard output An array of positive integers a1, a2, ..., an is given. Let us consider its arbitrary subarray al, al + 1..., ar

D. Powerful array time limit per test seconds memory limit per test megabytes input standard input output standard output An array of positive integers a1, a2, ..., an ..., ar,  ≤ l ≤ r ≤ n. For every positive integer s denote by Ks the number of occ

题目链接: D. Powerful array time limit per test 5 seconds memory limit per test 256 megabytes input standard input output standard output An array of positive integers a1, a2, ..., an is given. Let us consider its arbitrary subarray al, al + 1..., ar, wh

题目链接:http://codeforces.com/problemset/problem/86/D D. Powerful array time limit per test 5 seconds memory limit per test 256 megabytes input standard input output standard output An array of positive integers a1, a2, ..., an is given. Let us consider

题目链接 Powerful array 给你n个数,m次询问,Ks为区间内s的数目,求区间[L,R]之间所有Ks*Ks*s的和. $1<=n,m<=200000,   1<=s<=10^{6}$ 考虑莫队算法 把区间排序,然后让l和r分别询问即可. 根据排序的方式,l指针和r指针移动次数和大概是$n\sqrt{n}$ 级别的. #include <bits/stdc++.h> using namespace std; #define rep(i, a, b) for (

Powerful array 题意:求区间[l, r] 内的数的出现次数的平方 * 该数字. 题解:莫队离线操作, 然后加减位置的时候直接修改答案就好了. 这个题目中发现了一个很神奇的事情,本来数组开1e6大小就直接过了4100+ms, 想测试一下inline,顺手把空间砍成了刚好够用,然后跑的更慢了 4700+ms,删了inline之后T了,到现在也不知道发生了啥. 如果有大牛路过希望帮我看下,  CF run id  38822420(4100+) 38822607(4700+)388226

D. Powerful array time limit per test 5 seconds memory limit per test 256 megabytes input standard input output standard output An array of positive integers a1, a2, ..., an is given. Let us consider its arbitrary subarray al, al + 1..., ar, where 1 

Description An array of positive integers a1, a2, ..., an is given. Let us consider its arbitrary subarray al, al + 1..., ar, where 1 ≤ l ≤ r ≤ n. For every positive integer s denote by Ks the number of occurrences of s into the subarray. We call the

题意翻译 题意:给出一个n个数组成的数列a,有t次询问,每次询问为一个[l,r]的区间,求区间内每种数字出现次数的平方×数字的值 的和. 输入:第一行2个正整数n,t. 接下来一行n个正整数,表示数列a1​~an的值. 接下来t行,每行两个正整数l,r,为一次询问. 输出:t行,分别为每次询问的答案. 数据范围:1≤n,t≤2∗105,1≤ai≤106,1≤l,r≤n 题目描述 An array of positive integers a1,a2,...,an a_{1},a_{2},...,

Description An array of positive integers a1, a2, ..., an is given. Let us consider its arbitrary subarray al, al + 1..., ar, where 1 ≤ l ≤ r ≤ n. For every positive integer s denote by Ks the number of occurrences of s into the subarray. We call the

vjudge上莫队专题 真的是要吐槽自己(自己的莫队手残写了2个bug) s=sqrt(n) 是元素的个数而不是询问的个数(之所以是sqrt(n)使得左端点每个块左端点的范围嘴都是sqrt(n)) 在重载<是将q[i].l/s<q2[i].l/s 写成q1[i].l<s<q2[i].l<s 导致一下午都在调bug疯了 Sona 比小Z的袜子简单,直接维护区间频度的^3 需要离散化(离散化化的标号应提前准备好,如果用时在二分查找会增加复杂度) 还有比较坑的一点是多case,但样

题目描述: Problem Description An array of positive integers a1, a2, ..., an is given. Let us consider its arbitrary subarray al, al + 1..., ar, where 1 ≤ l ≤ r ≤ n. For every positive integer s denote by Ks the number of occurrences of s into the subarra

An array of positive integers a1, a2, ..., an is given. Let us consider its arbitrary subarray al, al + 1..., ar, where 1 ≤ l ≤ r ≤ n. For every positive integer s denote by Ks the number of occurrences of s into the subarray. We call the power of th

An array of positive integers a1, a2, ..., an is given. Let us consider its arbitrary subarray al, al + 1..., ar, where 1 ≤ l ≤ r ≤ n. For every positive integer s denote by Ks the number of occurrences of s into the subarray. We call the power of th

和BZOJ2038差不多..复习一下. #include<cstdio> #include<cmath> #include<algorithm> using namespace std; int block; struct Query{ int i,l,r; bool operator<(const Query &q)const{ if(l/block==q.l/block) return r<q.r; return l/block<q.l/b

给定一个数列:A1, A2,……,An,定义Ks为区间(l,r)中s出现的次数. t个查询,每个查询l,r,对区间内所有a[i],求sigma(K^2*a[i]) 离线+分块 将n个数分成sqrt(n)块. 对所有询问进行排序,排序标准: 1. Q[i].left /block_size < Q[j].left / block_size (块号优先排序) 2. 如果1相同,则 Q[i].right < Q[j].right (按照查询的右边界排序) 问题求解: 从上一个查询后的结果推出当前查询