codeforces 721B B. Passwords(贪心)
题目链接:
2 seconds
256 megabytes
standard input
standard output
Vanya is managed to enter his favourite site Codehorses. Vanya uses n distinct passwords for sites at all, however he can't remember which one exactly he specified during Codehorses registration.
Vanya will enter passwords in order of non-decreasing their lengths, and he will enter passwords of same length in arbitrary order. Just when Vanya will have entered the correct password, he is immediately authorized on the site. Vanya will not enter any password twice.
Entering any passwords takes one second for Vanya. But if Vanya will enter wrong password k times, then he is able to make the next try only 5 seconds after that. Vanya makes each try immediately, that is, at each moment when Vanya is able to enter password, he is doing that.
Determine how many seconds will Vanya need to enter Codehorses in the best case for him (if he spends minimum possible number of second) and in the worst case (if he spends maximum possible amount of seconds).
The first line of the input contains two integers n and k (1 ≤ n, k ≤ 100) — the number of Vanya's passwords and the number of failed tries, after which the access to the site is blocked for 5 seconds.
The next n lines contains passwords, one per line — pairwise distinct non-empty strings consisting of latin letters and digits. Each password length does not exceed 100 characters.
The last line of the input contains the Vanya's Codehorses password. It is guaranteed that the Vanya's Codehorses password is equal to some of his n passwords.
Consider the first sample case. As soon as all passwords have the same length, Vanya can enter the right password at the first try as well as at the last try. If he enters it at the first try, he spends exactly 1 second. Thus in the best case the answer is 1. If, at the other hand, he enters it at the last try, he enters another 4 passwords before. He spends 2 seconds to enter first 2 passwords, then he waits 5seconds as soon as he made 2 wrong tries. Then he spends 2 more seconds to enter 2 wrong passwords, again waits 5 seconds and, finally, enters the correct password spending 1 more second. In summary in the worst case he is able to be authorized in 15 seconds.
5 2
cba
abc
bb1
abC
ABC
abc
1 15
4 100
11
22
1
2
22
3 4 题意:
给了一堆密码,然后按长度从小到大开始试,问最少和最多的花费时间; 思路: 贪心; AC代码:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <bits/stdc++.h>
#include <stack>
#include <map> using namespace std; #define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
#define lson o<<1
#define rson o<<1|1
typedef long long LL; template<class T> void read(T&num) {
char CH; bool F=false;
for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
if(!p) { puts("0"); return; }
while(p) stk[++ tp] = p%10, p/=10;
while(tp) putchar(stk[tp--] + '0');
putchar('\n');
} const LL mod=1e9+7;
const double PI=acos(-1.0);
const LL inf=1e18;
const int N=1e6+10;
const int maxn=1e5+10;
const double eps=1e-12; int n,k,a[200];
string s[200],ans;
int main()
{
read(n);read(k);
for(int i=1;i<=n;i++){cin>>s[i],a[i]=s[i].length();}
cin>>ans;
int len=ans.length(),mi=0,eq=0;
for(int i=1;i<=n;i++)
{
if(a[i]<len)mi++;
else if(a[i]==len)eq++;
}
cout<<mi+mi/k*5+1<<" "<<mi+eq+(mi+eq-1)/k*5<<endl; return 0;
}
codeforces 721B B. Passwords(贪心)的更多相关文章
- 【37.21%】【codeforces 721B】Passwords
time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard o ...
- codeforces 704B - Ant Man 贪心
codeforces 704B - Ant Man 贪心 题意:n个点,每个点有5个值,每次从一个点跳到另一个点,向左跳:abs(b.x-a.x)+a.ll+b.rr 向右跳:abs(b.x-a.x) ...
- CodeForces - 50A Domino piling (贪心+递归)
CodeForces - 50A Domino piling (贪心+递归) 题意分析 奇数*偶数=偶数,如果两个都为奇数,最小的奇数-1递归求解,知道两个数都为1,返回0. 代码 #include ...
- Codeforces Round #374 (Div. 2) B. Passwords 贪心
B. Passwords 题目连接: http://codeforces.com/contest/721/problem/B Description Vanya is managed to enter ...
- CodeForces 721B Passwords (水题)
题意:给定 n 个密码,你要按长度不递减的顺序进行尝试,问你最多和最少试多少次可能找出密码,每尝试 k 次错误的,就要等5秒. 析:我们只要把长度全都统计下来,然后从1开始去找目标长度,最少的就是正好 ...
- Codeforces 161 B. Discounts (贪心)
题目链接:http://codeforces.com/contest/161/problem/B 题意: 有n个商品和k辆购物车,给出每个商品的价钱c和类别t(1表示凳子,2表示铅笔),如果一辆购物车 ...
- CodeForces 176A Trading Business 贪心
Trading Business 题目连接: http://codeforces.com/problemset/problem/176/A Description To get money for a ...
- Codeforces Gym 100803C Shopping 贪心
Shopping 题目连接: http://codeforces.com/gym/100803/attachments Description Your friend will enjoy shopp ...
- Codeforces 486C Palindrome Transformation(贪心)
题目链接:Codeforces 486C Palindrome Transformation 题目大意:给定一个字符串,长度N.指针位置P,问说最少花多少步将字符串变成回文串. 解题思路:事实上仅仅要 ...
随机推荐
- Spring RMI Example
一: 提供服务的远程一端 1-1. applicationContext.xml <?xml version="1.0" encoding="UTF-8" ...
- Hibernate的初步
1.简介 在java开发领域,基于数据库应用的设计与实现一直都是面向关系的,Hibernate对象/关系映射ORM框架的出现为java面向对象开发提供了易于使用的数据持久化解决方案. ORM介绍: ( ...
- mysql内存消耗分析
最近有些生产服务器老是mysql内存不停得往上涨,开发人员和维护反馈,用了不少的临时表,问题时常线上发生,测试又一直比较难重现. 经观察mysql内存的os占用趋势,发现从8:40开始,mysql内存 ...
- [iOS] 使用xib作为应用程序入口 with IDE
[iOS] 使用xib作为应用程序入口 with IDE 在「使用xib做为应用程序入口 with Code」这篇文章中,介绍了如何透过写Code的方式,来使用xib做为应用程序的入口.但其实在Xco ...
- javascript函数中的三个技巧【三】
技巧三: [函数绑定] 在javascript与DOM交互中经常需要使用函数绑定,定义一个函数然后将其绑定到特定DOM元素或集合的某个事件触发程序上,绑定函数经常和回调函数及事件处理程序一起使用,以便 ...
- Vue条件渲染
gitHub地址:https://github.com/lily1010/vue_learn/tree/master/lesson08 一 v-if显示单个元素 注意else只能跟在v-if或者v-s ...
- 编译hadoop eclipse的插件(hadoop1.0)
原创文章,转载请注明: 转载自工学1号馆 欢迎关注我的个人博客:www.wuyudong.com, 更多云计算与大数据的精彩文章 在hadoop-1.0中,不像0.20.2版本,有现成的eclipse ...
- C语言指针的长度和类型
本文地址:http://www.cnblogs.com/archimedes/p/point-length-type.html,转载请注明源地址. 如果考虑应用程序的兼容性和可移植性,指针的长度就是一 ...
- Python学习一入门
一.打印Hello和多行文本 print 打印 后跟单引号或者双引号 多行:3个单引号或者3个双引号 二.算术运算 2.1.加减乖法 默认1/2=0 如果需要小数运算,则需要一个运算术上加.或者.0 ...
- PL/SQL基础2(笔记)
1 第一个PL/SQL的程序 DECLARE BEGIN DBMS_OUTPUT.PUT_LINE('Hello World!'); END; / --2一个简单的PL/SQL程序 DECLARE v ...