zoukankan      html  css  js  c++  java
  • Palindrome(最长公共子序列)

    Time Limit: 3000MS   Memory Limit: 65536K
    Total Submissions: 48526   Accepted: 16674

    Description

    A palindrome is a symmetrical string, that is, a string read identically from left to right as well as from right to left. You are to write a program which, given a string, determines the minimal number of characters to be inserted into the string in order to obtain a palindrome. 

    As an example, by inserting 2 characters, the string "Ab3bd" can be transformed into a palindrome ("dAb3bAd" or "Adb3bdA"). However, inserting fewer than 2 characters does not produce a palindrome. 

    Input

    Your program is to read from standard input. The first line contains one integer: the length of the input string N, 3 <= N <= 5000. The second line contains one string with length N. The string is formed from uppercase letters from 'A' to 'Z', lowercase letters from 'a' to 'z' and digits from '0' to '9'. Uppercase and lowercase letters are to be considered distinct.

    Output

    Your program is to write to standard output. The first line contains one integer, which is the desired minimal number.

    Sample Input

    5
    Ab3bd

    Sample Output

    2

    题意:给一个字符串的长度和这个字符串s[],问将这个字符串变成回文串至少要加几个字符;

    思路:将这个字符串逆置s1[],求出s[]和s1[]的最长公共子序列的长度len,然后n-len即是要求的长度;

     1 #include<stdio.h>
     2 #include<string.h>
     3 short dp[5001][5001];
     4 int main()
     5 {
     6     int i,j,n;
     7     char s[5010],s1[5010];
     8     memset(dp,0,sizeof(dp));
     9 
    10     scanf("%d",&n);
    11     scanf("%s",s);
    12 
    13     for(int i = 0; i < n; i++)
    14         s1[i] = s[n-1-i];
    15     s1[n] = '';
    16 
    17     for(i = 0; i < n; i++)
    18     {
    19         for(j = 0; j < n; j++)
    20         {
    21             if(s[i] == s1[j])
    22                 dp[i+1][j+1] = dp[i][j]+1;
    23             else
    24             {
    25                 if(dp[i+1][j] > dp[i][j+1])
    26                     dp[i+1][j+1] = dp[i+1][j];
    27                 else dp[i+1][j+1] = dp[i][j+1];
    28             }
    29         }
    30     }
    31     int ans = n-dp[n][n];
    32     printf("%d
    ",ans);
    33     return 0;
    34 }
    View Code
  • 相关阅读:
    JAVA基础集合(二)
    JAVA基础XML的解析
    JAVA基础集合(一)
    JAVA基础网络编程
    JAVA基础File类
    JAVA基础IO流(二)
    JAVA基础JDBC二(常用的开源工具)
    JAVA基础异常
    JAVA基础JDBC(一)
    JAVA基础反射
  • 原文地址:https://www.cnblogs.com/LK1994/p/3290778.html
Copyright © 2011-2022 走看看