zoukankan      html  css  js  c++  java
  • 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


    Problem Description
    Setting password is very important, especially when you have so many "interesting" things in "F:TDDOWNLOAD".
    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.
     
    Input
    The first line has a number T (T <= 15) , indicating the number of test cases.
    For each test case, there is a single line with two integers M and N, as described above.(1 <= M <= 26, 1 <= N <= 105)
     
    Output
    For test case X, output "Case #X: " first, then output the best password.
     
    Sample Input
    2 2 2 2 3
     
    Sample Output
    Case #1: ab Case #2: aab
     
    Source
     
    Recommend
    liuyiding
     

    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

     1 /* ***********************************************
     2 Author        :kuangbin
     3 Created Time  :2013/9/14 星期六 13:21:24
     4 File Name     :2013成都网络赛1004.cpp
     5 ************************************************ */
     6 
     7 #pragma comment(linker, "/STACK:1024000000,1024000000")
     8 #include <stdio.h>
     9 #include <string.h>
    10 #include <iostream>
    11 #include <algorithm>
    12 #include <vector>
    13 #include <queue>
    14 #include <set>
    15 #include <map>
    16 #include <string>
    17 #include <math.h>
    18 #include <stdlib.h>
    19 #include <time.h>
    20 using namespace std;
    21 
    22 char str[10000];
    23 int main()
    24 {
    25     //freopen("in.txt","r",stdin);
    26     //freopen("out.txt","w",stdout);
    27     int T;
    28     int m,n;
    29     int iCase = 0;
    30     scanf("%d",&T);
    31     while(T--)
    32     {
    33         scanf("%d%d",&m,&n);
    34         iCase++;
    35         printf("Case #%d: ",iCase);
    36         if(m == 1)
    37         {
    38             for(int i = 0;i < n;i++)printf("a");
    39             printf("
    ");
    40             continue;
    41         }
    42         if(m >= 3)
    43         {
    44             int index = 0;
    45             for(int i = 0;i < n;i++)
    46             {
    47                 printf("%c",index + 'a');
    48                 index = (index + 1)%3;
    49             }
    50             printf("
    ");
    51             continue;
    52         }
    53         if(n == 1)printf("a");
    54         else if(n == 2)printf("ab");
    55         else if(n == 3)printf("aab");
    56         else if(n == 4)printf("aabb");
    57         else if(n == 5)printf("aaaba");
    58         else if(n == 6)printf("aaabab");
    59         else if(n == 7)printf("aaababb");
    60         else if(n == 8)printf("aaababbb");
    61         else if(n == 9)printf("aaaababba");
    62         else
    63         {
    64             printf("aaaa");
    65             n -= 4;
    66             while(n >= 6)
    67             {
    68                 printf("babbaa");
    69                 n -= 6;
    70             }
    71             if(n == 1)printf("a");
    72             else if(n == 2)printf("aa");
    73             else if(n == 3)printf("bab");
    74             else if(n == 4)printf("babb");
    75             else if(n == 5)printf("babba");
    76         }
    77         printf("
    ");
    78     }
    79     return 0;
    80 }
  • 相关阅读:
    利用哈希map快速判断两个数组的交集
    TCP协议中的三次握手和四次挥手(图解)-转
    PC,移动端H5实现实现小球加入购物车效果
    HQL和SQL的区别
    Java泛型详解,通俗易懂只需5分钟
    经典的 Fork 炸弹解析
    Java并发之AQS详解
    Java不可重入锁和可重入锁的简单理解
    Codeforces 1215F. Radio Stations
    Codeforces 1215E. Marbles
  • 原文地址:https://www.cnblogs.com/kuangbin/p/3321994.html
Copyright © 2011-2022 走看看