Codeforces —— Palindromic characteristics(835D)

input
abba
output
6 1 0 0 

input
abacaba
output
12 4 1 0 0 0 0 

题意：

1-palindromes:该字符串为一个回文字符串
k-palindromes满足：
1.该字符串为一个回文字符串
2.其左边以及右边是一个(k-1)-palindromes
询问给定字符串中有几个i-palindromes;

思路：

考虑区间DP，dp[i][j] —— 字符串i~j是最多是几阶回文
状态转移方程：dp[i][j] = dp[i][i+len/2-1] + 1;

代码

#include<unordered_map>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<queue>
#include<stack>
#include<cmath>
#include<map>
#include<set>
#define Buff ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define rush() int Case = 0; int T; scanf("%d", &T);  while(T--)
#define rep(i, a, b) for(int i = a; i <= b; i ++)
#define per(i, a, b) for(int i = a; i >= b; i --)
#define reps(i, a, b) for(int i = a; b; i ++)
#define clc(a, b) memset(a, b, sizeof(a))
#define Buff ios::sync_with_stdio(false)
#define readl(a) scanf("%lld", &a)
#define readd(a) scanf("%lf", &a)
#define readc(a) scanf("%c", &a)
#define reads(a) scanf("%s", a)
#define read(a) scanf("%d", &a)
#define lowbit(n) (n&(-n))
#define pb push_back
#define sqr(x) x*x
#define rs x<<1|1
#define y second
#define ls x<<1
#define x first
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int>PII;
const int mod = 1e9+7;
const double eps = 1e-6;
const int N = 5e3+7;
int res[N], f[N][N];
char s[N];
int main()
{
	Buff;
	cin >> s+1;
	int n = strlen(s+1);
	res[1] = n;
	rep(i, 1, n)	f[i][i] = 1;
	for(int len = 2; len <= n; len ++)
		for(int l = 1; l + len - 1 <= n; l ++)
		{
			int r = l + len - 1;
			if(s[l] != s[r] || (l+1 <= r-1) && !f[l+1][r-1])	f[l][r] = 0;
			else												f[l][r] += f[l][l+len/2-1] + 1;
			res[f[l][r]] ++;
		}
	per(i, n-1, 1)	res[i] += res[i+1];
	rep(i, 1, n)	cout << res[i] <<" ";puts("");
	return 0;
}