C. DZY Loves Colors time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output DZY loves colors, and he enjoys painting. On a colorful day, DZY gets a colorful ribbon, which consists of n units (they…

题目 Source http://codeforces.com/problemset/problem/444/C Description DZY loves colors, and he enjoys painting. On a colorful day, DZY gets a colorful ribbon, which consists of n units (they are numbered from 1 to n from left to right). The color of t…

题目链接: http://codeforces.com/problemset/problem/444/C J. DZY Loves Colors time limit per test:2 secondsmemory limit per test:256 megabytes 问题描述 DZY loves colors, and he enjoys painting. On a colorful day, DZY gets a colorful ribbon, which consists of…

题目大意:Codeforces 444C DZY Loves Colors 题目大意:两种操作,1是改动区间上l到r上面德值为x,2是询问l到r区间总的改动值. 解题思路:线段树模板题. #include <cstdio> #include <cstring> #include <cstdlib> #include <algorithm> using namespace std; const int maxn = 5*1e5; typedef long lo…

DZY loves colors, and he enjoys painting. On a colorful day, DZY gets a colorful ribbon, which consists of n units (they are numbered from 1 to n from left to right). The color of thei-th unit of the ribbon is i at first. It is colorful enough, but w…

C - DZY Loves Colors 思路: 分块,复杂度有点玄学,和普通分块不同的是在这个块被一次染色的时候暴力染整个块. 代码: #pragma GCC optimize(2) #pragma GCC optimize(3) #pragma GCC optimize(4) #include<bits/stdc++.h> using namespace std; #define fi first #define se second #define pi acos(-1.0) #defin…

C. DZY Loves Colors 题目连接: http://codeforces.com/contest/444/problem/C Description DZY loves colors, and he enjoys painting. On a colorful day, DZY gets a colorful ribbon, which consists of n units (they are numbered from 1 to n from left to right). T…

DZY Loves Colors Time Limit: 2000ms Memory Limit: 262144KB This problem will be judged on CodeForces. Original ID: 445E64-bit integer IO format: %I64d      Java class name: (Any) DZY loves colors, and he enjoys painting. On a colorful day, DZY gets a…

题意:输入n, m ; 有n给位置, 初始时第i个位置的color为i, colorfulness为0.      有m次操作,一种是把成段的区域color更新为x, 对于更新的区域,每个位置(令第i个位置未更新前颜色为color[i])的colorfulness增加|color[i] -x|;      另一种操作是询问一段区间[L,R]输出该区间的colorfulness总和.   代码实现 #include <iostream> #include <cstdio> #incl…

考试完之后打的第一场CF,异常惨烈呀,又只做出了一题了.A题呆滞的看了很久,领悟到了出题者的暗示,应该就是两个点的时候最大吧,不然的话这题肯定特别难敲,YY一发交上去然后就过了.然后就在不停地YY B题,赛后听了英姐的答案,看了题解,发现其实自己是捕捉到了正确的解题思路的,但是因为不知道怎么算出期望的复杂度,因而就没敢敲,哪里会想到算复杂度会算期望的情况的呢- -0 然后就来说下坑爹的C题了,比赛的时候看了觉得是线段树,它支持段更,但是每次段更的时候对于每个点,更新的差值|x-y|要被记录下来,…