Codeforces 159D palindrome Pairs

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

Main topic:

Gives a string that takes the logarithm of two palindrome substrings that are not covered by each other in this string.

Thought: Num[i] represents the left point at the I position of the palindrome string number, and then the tree-like array maintenance sum[i], on behalf of the Palindrome string right end of the palindrome string of less than equals I, total complexity: O (n^2)

1#include <cstdio>2#include <cmath>3#include <algorithm>4#include <cstring>5#include <iostream>6 #definell Long Long7ll c[200005],num[200005];8 intpd[2005][2005],n;9 Chars[200005];Ten intLowbit (intx) { One     returnx& (-x); A } - intRead () { -     intt=0, f=1;CharCh=GetChar (); the      while(ch<'0'|| Ch>'9'){if(ch=='-') f=-1; ch=GetChar ();} -      while('0'<=ch&&ch<='9') {t=t*Ten+ch-'0'; ch=GetChar ();} -     returnt*F; - } + voidAddintx) { -      for(inti=x;i<=n;i+=lowbit (i)) { +c[i]++; A     } at } - intAskintx) { -     intres=0; -      for(inti=x;i;i-=lowbit (i)) { -res+=C[i]; -     } in     returnRes; - } to intMain () { +scanf"%s", s+1); -N=strlen (s+1); the      for(intI=1; i<=n;i++) *pd[i][i]=1, add (i), num[i]++; $      for(intI=1; i<n;i++)Panax Notoginseng      if(s[i]==s[i+1]) pd[i][i+1]++,add (i+1), num[i]++; -      for(intlen=3; len<=n;len++) the       for(intI=1; i+len-1<=n;i++){ +         intj=i+len-1; A         if(pd[i+1][j-1]&AMP;&AMP;S[I]==S[J]) pd[i][j]=1, add (j), num[i]++; the      } +ll ans=0;  -      for(intI=1; i<=n;i++) $Ans+=ask (I-1)*( ll) (Num[i]); $printf"%i64d\n", ans); -     return 0; -}

Codeforces 159D palindrome Pairs