zoukankan      html  css  js  c++  java
  • Palindrome

    Palindrome

    Time Limit : 4000/2000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)
    Total Submission(s) : 88   Accepted Submission(s) : 30
    Problem 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
     代码:
     1 #include<stdio.h>
     2 #include<string.h>
     3 const int MAXN=5010;
     4 const int INF=0xfffffff;
     5 #define MAX(x,y) (x>y?x:y)
     6 int dp[MAXN][2];
     7 char a[MAXN],b[MAXN];
     8 int main(){
     9     int t,i,j;
    10     while(~scanf("%d",&t)){
    11         scanf("%s",a+1);
    12         //printf("%d
    ",t);
    13         for(i=t,j=1;i>0;i--,j++)b[j]=a[i];
    14         b[j]='';
    15         //printf("%s
    ",b+1);
    16         memset(dp,0,sizeof(dp));
    17         dp[0][0]=0;
    18         for(i=1;i<=t;i++){
    19             for(j=1;j<=t;j++){
    20                 if(a[i]==b[j])dp[j][i%2]=dp[j-1][(i-1)%2]+1;
    21                 else dp[j][i%2]=MAX(dp[j-1][i%2],dp[j][(i-1)%2]);
    22             }
    23         }
    24         //printf("%d
    ",dp[t][t%2]);
    25         printf("%d
    ",t-dp[t][t%2]);
    26     }
    27 return 0;}

    题解;还可以是%2,%3,%4,........

  • 相关阅读:
    计算机网络常见面试题
    字节跳动2022秋招提前批来了!!!
    IBM Watson Studio
    Git提交GitHub
    python获取股票和基金等数据
    Cloud Foundry
    微软亚洲研究院的NLP一例
    Streamlit的学习小记
    在线学习云技术相关等
    IBM云部署相关
  • 原文地址:https://www.cnblogs.com/handsomecui/p/4719360.html
Copyright © 2011-2022 走看看