C. Dasha and Password time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output After overcoming the stairs Dasha came to classes. She needed to write a password to begin her classes. The password i…

C. Dasha and Password time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output After overcoming the stairs Dasha came to classes. She needed to write a password to begin her classes. The password i…

C. Dasha and Password 题目连接: http://codeforces.com/contest/761/problem/C Description After overcoming the stairs Dasha came to classes. She needed to write a password to begin her classes. The password is a string of length n which satisfies the follo…

C. Dasha and Password time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output After overcoming the stairs Dasha came to classes. She needed to write a password to begin her classes. The password i…

题目链接:http://codeforces.com/problemset/problem/761/C C. Dasha and Password time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output After overcoming the stairs Dasha came to classes. She needed to w…

http://codeforces.com/contest/761/problem/C 题意:给出n个串,每个串的初始光标都位于0(列)处,怎样移动光标能够在凑出密码(每个串的光标位置表示一个密码的字符,密码至少包含3种字符:数字,小写字母,特殊符号)的情况下使得移动的光标步数最小. 思路: 因为每个串只提供一个密码,所以我们先预处理计算出每个字符串3种字符的最少移动步数. 然后接下三重循环枚举,分别表示数字,小写字母,特殊符号由第i,j,k行提供. #include<iostream> #i…

纪念死去的智商(虽然本来就没有吧……) 三重循环枚举将哪三个fix string作为数字.字母和符号位.记下最小的值就行了. 预处理之后这个做法应该是O(n^3)的,当然完全足够.不预处理是O(n^3*m)的,也够. 我写了一个O(n^2+n*m)的分类讨论贪心做法……蜜汁错,容我查一下. 现在查出个错,交了一下在in queue……容我明天看看对不对. 如果对的话,这题的加强版今年留作趣味赛题吧,嘿嘿嘿…… UPDATE:果然过了……我就是傻逼 #include<cstdio> #inclu…

E. Dasha and Puzzle time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Dasha decided to have a rest after solving the problem. She had been ready to start her favourite activity — origami, bu…

题目链接 Dasha and Password 题目保证一定有解. 考虑到最多只有两行的指针需要移动,那么直接预处理出该行移动到字母数字或特殊符号的最小花费. 然后O(N^3)枚举求最小值即可. 时间复杂度O(N*M+N^3) #include <bits/stdc++.h> using namespace std; #define REP(i,n) for(int i(0); i < (n); ++i) #define rep(i,a,b) for(int i(a); i <=…

time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard output After overcoming the stairs Dasha came to classes. She needed to write a password to begin her classes. The password is a string of length n whi…