hduoj1544(判断连续回文)

Palindromes

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 1509 Accepted Submission(s): 712

Problem Description

A regular palindrome(回文) is a string of numbers or letters that is the same forward as backward. For example, the string “ABCDEDCBA” is a palindrome because it is the same when the string is read from left to right as when the string is read from right to left.

Now give you a string S, you should count how many palindromes in any consecutive substring of S.

Input

There are several test cases in the input. Each case contains a non-empty string which has no more than 5000 characters.

Proceed to the end of file.

Output

A single line with the number of palindrome substrings for each case.

Sample Input

aba

aa

Sample Output

4

3

题解：从某一个开始，然后分偶数和奇数向左向右扩展，遇到不相等就退出，用二维dp会超时。

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
char str[5005];
bool dp[5005][5005];
int main(){
int len,i,j,cnt,l,r;
while(scanf("%s",str)!=EOF){

len=strlen(str);
cnt=len;
for(i = 0 ; i < len ; i++){
l=i-1;
r=i+1;
while(l>=0&&r<len&&str[l]==str[r]){
l--;r++;
cnt++;
}
l=i;r=i+1;
while(l>=0&&r<len&&str[l]==str[r]){
l--;r++;
cnt++;
}
}
printf("%d\n",cnt);
//cout<<cnt<<endl;
}
return 0;
}