D. Equalize Them All time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output You are given an array aa consisting of nn integers. You can perform the following operations arbitrary number of times…

题目连接:Equalize the Remainders 题意:n个数字,对m取余有m种情况,使得每种情况的个数都为n/m个(保证n%m=0),最少需要操作多少次? 每次操作可以把某个数字+1.输出最少操作次数,和操作后的序列(可以输出任意一种). 题解:用一个set来维护所有余数x(当前余数为x的数个数没凑够n/m个),对于每个数假设这个数的余数为t,当余数为t的数个数没凑够n/m时那这个数就不需要改变,如果已经凑够了,那就在set中找到第一个大于等于t的数(注意这里t可能比set中最大数的还…

modify 改变 C. Equalize time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output You are given two binary strings aa and bb of the same length. You can perform the following two operations on the stri…

D. Equalize the Remainders time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output You are given an array consisting of nn integers a1,a2,…,ana1,a2,…,an, and a positive integer mm. It is guarantee…

D. Equalize Them All time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output You are given an array aa consisting of nn integers. You can perform the following operations arbitrary number of times…

D. Equalize Them All You are given an array aa consisting of nn integers. You can perform the following operations arbitrary number of times (possibly, zero): Choose a pair of indices (i,j)(i,j) such that |i−j|=1|i−j|=1 (indices ii and jj are adjacen…

洛谷 CF1234A Equalize Prices Again 洛谷传送门 题目描述 You are both a shop keeper and a shop assistant at a small nearby shop. You have nn goods, the ii -th good costs a_i*a**i* coins. You got tired of remembering the price of each product when customers ask fo…

D. Equalize the Remainders set的学习::https://blog.csdn.net/byn12345/article/details/79523516 注意set的end()和rbegin()的区别. end()是指向最后一个元素的下一个,rbegin()是指向最后一个元素. 题目大意:给你一个n长度的数组,给一个模数m,问对m取模,余数从0到m-1的每一种都是n/m 保证m一定是n的除数. 每一个操作对数字+1,问最少的操作满足题目,输出操作之后的数组. 这个用s…

Codeforce 999 D. Equalize the Remainders 解析(思維) 今天我們來看看CF999D 題目連結 題目 略,請直接看原題 前言 感覺要搞個類似\(stack\)的東西來儲存下一個沒滿的\(\mod m\)是哪一個才能避免\(O(m^2)\)的複雜度,沒想到反過來想,儲存前一個滿出來的是什麼就可以了. @copyright petjelinux 版權所有 觀看更多正版原始文章請至petjelinux的blog 想法 首先可能會想到,先把每個\(mod\)值都儲存…

原题 题目大意: 给你两个长度都为\(n\)的的\(01\)串\(a,b\),现在你可以对\(a\)串进行如下两种操作: 1.交换位置\(i\)和位置\(j\),代价为\(|i-j|\) 2.反转位置\(i\) \((0->1, 1->0)\), 代价为\(1\) 问你将\(a\)变成\(b\)的最小代价是多少. 乍一看,像是dp.但是,仔细想一想,会发现只有在\(i\)和\(j\)的距离为\(1\),且\(a[i] \neq b[i] \ and \ a[j] \neq b[i] \ and…