HDU 4731 Minimum palindrome (2013成都网络赛,找规律构造)
Minimum palindrome
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 260 Accepted Submission(s): 127
We define the safety of a password by a value. First, we find all the substrings of the password. Then we calculate the maximum length of those substrings which, at the meantime, is a palindrome.
A palindrome is a string that will be the same when writing backwards. For example, aba, abba,abcba are all palindromes, but abcab, abab are not.
A substring of S is a continous string cut from S. bcd, cd are the substrings of abcde, but acd,ce are not. Note that abcde is also the substring of abcde.
The smaller the value is, the safer the password will be.
You want to set your password using the first M letters from the alphabet, and its length should be N. Output a password with the smallest value. If there are multiple solutions, output the lexicographically smallest one.
All the letters are lowercase.
For each test case, there is a single line with two integers M and N, as described above.(1 <= M <= 26, 1 <= N <= 105)
2 2
2 3
Case #2: aab
找规律。
当m=1时,直接输出n个a
当m>=3时,输出abcabcabc....
当m=2时,先暴力打出1~20的表。然后找规律发现循环
/* ***********************************************
Author :kuangbin
Created Time :2013/9/14 星期六 13:21:24
File Name :2013成都网络赛\1004.cpp
************************************************ */ #pragma comment(linker, "/STACK:1024000000,1024000000")
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std; char str[];
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int T;
int m,n;
int iCase = ;
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&m,&n);
iCase++;
printf("Case #%d: ",iCase);
if(m == )
{
for(int i = ;i < n;i++)printf("a");
printf("\n");
continue;
}
if(m >= )
{
int index = ;
for(int i = ;i < n;i++)
{
printf("%c",index + 'a');
index = (index + )%;
}
printf("\n");
continue;
}
if(n == )printf("a");
else if(n == )printf("ab");
else if(n == )printf("aab");
else if(n == )printf("aabb");
else if(n == )printf("aaaba");
else if(n == )printf("aaabab");
else if(n == )printf("aaababb");
else if(n == )printf("aaababbb");
else if(n == )printf("aaaababba");
else
{
printf("aaaa");
n -= ;
while(n >= )
{
printf("babbaa");
n -= ;
}
if(n == )printf("a");
else if(n == )printf("aa");
else if(n == )printf("bab");
else if(n == )printf("babb");
else if(n == )printf("babba");
}
printf("\n");
}
return ;
}
HDU 4731 Minimum palindrome (2013成都网络赛,找规律构造)的更多相关文章
- HDU 4733 G(x) (2013成都网络赛,递推)
G(x) Time Limit: 2000/500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submiss ...
- HDU 4731 Minimum palindrome 2013 ACM/ICPC 成都网络赛
传送门:http://acm.hdu.edu.cn/showproblem.php?pid=4731 题解:规律题,我们可以发现当m大于等于3时,abcabcabc……这个串的回文为1,并且字典数最小 ...
- HDU 4737 A Bit Fun (2013成都网络赛)
A Bit Fun Time Limit: 5000/2500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total S ...
- HDU 4734 F(x) (2013成都网络赛,数位DP)
F(x) Time Limit: 1000/500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submiss ...
- HDU 4730 We Love MOE Girls (2013成都网络赛,签到水题)
We Love MOE Girls Time Limit: 1000/500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) ...
- HDU 4768 Flyer (2013长春网络赛1010题,二分)
Flyer Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submi ...
- HDU 4747 Mex (2013杭州网络赛1010题,线段树)
Mex Time Limit: 15000/5000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)Total Submis ...
- HDU 4731 Minimum palindrome 打表找规律
http://acm.hdu.edu.cn/showproblem.php?pid=4731 就做了两道...也就这题还能发博客了...虽然也是水题 先暴力DFS打表找规律...发现4个一组循环节.. ...
- hdu 4731 2013成都赛区网络赛 找规律
题意:找字串中最长回文串的最小值的串 m=2的时候暴力打表找规律,打表可以用二进制枚举
随机推荐
- shell 判断路径
判断路径 ];then echo "找到了123" if [ -d /root/Desktop/text ] then echo "找到了text" else ...
- CF11D A Simple Task(状压DP)
\(solution:\) 思路大家应该都懂: 状压DP:\(f[i][j]\),其中 \(i\) 这一维是需要状压的,用来记录19个节点每一个是否已经走过(走过为 \(1\) ,没走为 \(0\) ...
- 2016最新的中国省市区三级数据库表.sql mssql
/****** Object: Table [dbo].[t_Area] Script Date: 09/10/2016 09:35:46 ******/ SET ANSI_NULLS ON GO S ...
- 记关于webpack4下css提取打包去重复的那些事
注意使用vue-cli3(webpack4),默认小于30k不会抽取为公共文件,包括css和js,已测试 经过2天的填坑,现在终于有点成果 环境webpack4.6 + html-webpack-pl ...
- 【干货】linux系统信息收集 ----检测是否被恶意程序执行了危险性命令
这些实战完全可以练习以下命令,已经找到需要观察的交互点,真实工作的时候,把数据都导入到自己U盘或者工作站内. 在kali 或者centos下训练都一样,关于kali教学,这里推荐掌控安全团队的课程:掌 ...
- 字符串格式化(百分号&format)
字符串格式化 Python的字符串格式化有两种方式: 百分号方式.format方式 百分号方式: %[(name)][flags][width].[precision]typecode [ ]:表示 ...
- OpenWRT开发之——对C++的支持(解决库依赖问题)【转】
转自:https://my.oschina.net/hevakelcj/blog/411944 摘要: 本文尝试用C++来开发一个cpp-demo包 遇到打包库依赖的问题,分析打包过程并解决了这个问题 ...
- nginx tomcat 自动部署python脚本【转】
#!/usr/bin/env python #--coding:utf8-- import sys,subprocess,os,datetime,paramiko,re local_path='/ho ...
- C# UDP广播消息
首先是发送端: /// <summary> /// 发送UDP消息 /// </summary> /// <param name="msg">消 ...
- logback.xml 模板
ssm模板 <?xml version="1.0" encoding="UTF-8"?> <!--configuration 根节点,包含下 ...