Codeforces 245H Queries for Number of Palindromes

http://codeforces.com/contest/245/problem/H

题意:给定一个字符串，每次给个区间，求区间内有几个回文串(n<=5000)

思路:设定pd[i][j]代表i~j这部分是不是一个回文串，这个可以n^2预处理

然后设定f[i][j]代表i~j区间有多少个回文串，由于满足区间加减，因此有

f[i][j]=f[i+1][j]+f[i][j-1]-f[i+1][j-1]+pd[i][j]

 #include<cstdio>
 #include<cmath>
 #include<algorithm>
 #include<cstring>
 #include<iostream>
 int pd[][],f[][],n;
 char s[];
 int read(){
     int t=,f=;char ch=getchar();
     while (ch<''||ch>''){if (ch=='-') f=-;ch=getchar();}
     while (''<=ch&&ch<=''){t=t*+ch-'';ch=getchar();}
     return t*f;
 }
 void init(){
     for (int i=;i<=n;i++)
      pd[i][i]=;
     for (int len=;len<=n;len++)
      for (int i=;i+len-<=n;i++){
             int j=i+len-;
             if (i==j-&&s[i]==s[j]) pd[i][j]=;
             else
             if (s[i]==s[j]) pd[i][j]|=pd[i+][j-];
     }
     for (int i=;i<=n;i++)
      f[i][i]=;
     for (int i=;i<n;i++)
      if (s[i]==s[i+])
       f[i][i+]=;
      else
       f[i][i+]=;
     for (int len=;len<=n;len++)
      for (int i=;i+len-<=n;i++){
         int j=i+len-;
         f[i][j]=f[i+][j]+f[i][j-]-f[i+][j-]+pd[i][j];
      }
 }
 int main(){
     scanf("%s",s+);
     n=strlen(s+);
     init();
     int T=read();
     while (T--){
         int l=read(),r=read();
         printf("%d\n",f[l][r]);
     }
     return ;
 }