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:
    c语言几个字符串处理函数的简单实现
    各种类型排序的实现及比较
    随机洗牌算法Knuth Shuffle和错排公式
    两个栈实现队列
    面试杂题
    面试题——栈的压入、弹出顺序
    Codeforces 455A. Boredom
    PAT A1049. Counting Ones (30)
    Codeforces 895B. XK Segments
    Codeforces 282C. XOR and OR
  • 原文地址:https://www.cnblogs.com/handsomecui/p/4719360.html
Copyright © 2011-2022 走看看