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[200005],num[200005];
int pd[2005][2005],n;
char s[200005];
int lowbit(int x){
    return x&(-x);
}
int read(){
    int t=0,f=1;char ch=getchar();
    while (ch<'0'||ch>'9'){if (ch=='-') f=-1;ch=getchar();}
    while ('0'<=ch&&ch<='9'){t=t*10+ch-'0';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=0;
    for (int i=x;i;i-=lowbit(i)){
        res+=c[i];
    }
    return res;
}
int main(){
    scanf("%s",s+1);
    n=strlen(s+1);
    for (int i=1;i<=n;i++)
     pd[i][i]=1,add(i),num[i]++;
    for (int i=1;i<n;i++)
     if (s[i]==s[i+1]) pd[i][i+1]++,add(i+1),num[i]++;
    for (int len=3;len<=n;len++)
     for (int i=1;i+len-1<=n;i++){
        int j=i+len-1;
        if (pd[i+1][j-1]&&s[i]==s[j]) pd[i][j]=1,add(j),num[i]++;
     }
    ll ans=0; 
    for (int i=1;i<=n;i++)
      ans+=ask(i-1)*((ll)(num[i]));
    printf("%I64d\n",ans);
    return 0;
}