Codeforces Round #427 (Div. 2) D.Palindromic characteristics

题目链接
题意是这样的:定义了一种k阶回文串，1-阶是普通的回文串，k-阶回文串的前半串和后半串必须是相等的(不是回文，例如abcabc,如果长度奇数中间的不管他)然后其前半串必须是k-1阶回文串，后半串必须是k-1阶回文串。
然后给一个串，问这个串有多少子串是i阶回文串(i从1到n，依次输出)
分析一下可以看出如果一个串是k阶回文串，那么他必然是k-1阶回文串，于是我们可以枚举该串的子串记录他最大能到达的阶数，然后通过区间dp转移一下。由于阶数的每次增长都必然伴随着串长的加倍，所以k最大只有logn个。直接区间dp复杂度o(n² logn)

#include<algorithm>#include<cstring>#include<string>#include<set>#include<map>#include<time.h>#include<cstdio>#include<vector>#include<list>#include<stack>#include<queue>#include<iostream>#include<stdlib.h>using namespace std;#define  LONG long longconst int   INF=0x3f3f3f3f;const LONG  MOD=1e9+ 7;const double PI=acos(-1.0);#define clrI(x) memset(x,-1,sizeof(x))#define clr0(x) memset(x,0,sizeof x)#define clr1(x) memset(x,INF,sizeof x)#define clr2(x) memset(x,-INF,sizeof x)#define EPS 1e-10#define lson  l , mid , rt<< 1#define rson  mid + 1 ,r , (rt<<1)+1#define root 1, n , 1char str[5010] ;int ans [5010] ;int dp[5010][5010] ;int match[5010][5010] ;int main(){    while(~scanf("%s",str+1))    {        clr0(ans ) ;        clr0(dp) ;        str[0] = '#' ;        int len = strlen(str ) ;        len --;        for(int i = 1 ; i <= len+5 ; ++i)            dp[i][i-1] = 1 ;        for(int i = 2 ; i <= len+5 ; ++i)            dp[i][i-2] = 1 ;            //for (int i=1;i<=len;i++)     match[i][i]=1;     for (int l=1;l<=len;l++)      for (int i=1;i+l<=len;i++)        if  ( str[i] == str[i+l])          {              if (l==1)              match[i][i+l] = 1;              else              match[i][i+l] |= match[i+1][i+l-1];          }//        for(int l = 1 ; l <= len ; ++ l)        {            for(int i = 1 ; i <= len ; ++ i)            {                int j = i + l -1;                if( j > len ) break ;                if(str[i] == str[j]) if(dp[i+1][j-1]) dp[i][j] = 1;                int mid = l/2  ;                if(l <= 1) continue ;                if(l&1)                {                    if(dp[i][i+mid-1] == dp[i+mid+1][j] )                    {                        if( match[i][j] )                        {                            if(str[i] == str[j]) dp[i][j] = max(dp[i][j] , dp[i][i+mid-1] + 1) ;                        }                    }                }                else                {                    if(dp[i][i+mid-1] == dp[i+mid][j])                    {                            if(str[i] == str[j] && match[i][j] )                                dp[i][j] = max(dp[i][j] , dp[i][i+mid-1] + 1 );                    }                }            }        }        for(int i = 1 ;i<= len ; ++ i)        {            for(int j = i ; j<= len ; ++ j)                for(int k = 1 ;k<= dp[i][j] ; ++ k)                    ans[k] ++ ;        }        for(int i = 1;i <len ; ++ i)printf("%d ",ans[i]) ;printf("%d\n",ans[len]) ;    }}