Codeforces 159D Palindrome pairs

http://codeforces.com/problemset/problem/159/D

题目大意：

给出一个字符串，求取这个字符串中互相不覆盖的两个回文子串的对数。

思路:num[i]代表左端点在i这个位置的回文串个数，然后用树状数组维护sum[i],代表回文串右端点小于等于i的回文串数，总复杂度:O(n^2)

 #include<cstdio>
 #include<cmath>
 #include<algorithm>
 #include<cstring>
 #include<iostream>
 #define ll long long
 ll c[],num[];
 int pd[][],n;
 char s[];
 int lowbit(int x){
     return x&(-x);
 }
 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 add(int x){
     for (int i=x;i<=n;i+=lowbit(i)){
         c[i]++;
     }
 }
 int ask(int x){
     int res=;
     for (int i=x;i;i-=lowbit(i)){
         res+=c[i];
     }
     return res;
 }
 int main(){
     scanf("%s",s+);
     n=strlen(s+);
     for (int i=;i<=n;i++)
      pd[i][i]=,add(i),num[i]++;
     for (int i=;i<n;i++)
      if (s[i]==s[i+]) pd[i][i+]++,add(i+),num[i]++;
     for (int len=;len<=n;len++)
      for (int i=;i+len-<=n;i++){
         int j=i+len-;
         if (pd[i+][j-]&&s[i]==s[j]) pd[i][j]=,add(j),num[i]++;
      }
     ll ans=;
     for (int i=;i<=n;i++)
       ans+=ask(i-)*((ll)(num[i]));
     printf("%I64d\n",ans);
     return ;
 }