Educational Codeforces Round 41 (Rated for Div. 2) F. k-substrings

F. k-substrings

time limit per test

4 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a string s consisting of n lowercase Latin letters.

Let's denote k-substring of s as a string subsk = sksk + 1..sn + 1 - k. Obviously, subs1 = s, and there are exactly  such substrings.

Let's call some string t an odd proper suprefix of a string T iff the following conditions are met:

• |T| > |t|;
• |t| is an odd number;
• t is simultaneously a prefix and a suffix of T.

For evey k-substring () of s you have to calculate the maximum length of its odd proper suprefix.

Input

The first line contains one integer n (2 ≤ n ≤ 106) — the length s.

The second line contains the string s consisting of n lowercase Latin letters.

Output

Print  integers. i-th of them should be equal to maximum length of an odd proper suprefix of i-substring of s (or  - 1, if there is no such string that is an odd proper suprefix of i-substring).

Examples

input

Copy

15
bcabcabcabcabca

output

Copy

9 7 5 3 1 -1 -1 -1

input

Copy

24
abaaabaaaabaaabaaaabaaab

output

Copy

15 13 11 9 7 5 3 1 1 -1 -1 1

input

Copy

19
cabcabbcabcabbcabca

output

Copy

5 3 1 -1 -1 1 1 -1 -1 -1

Note

The answer for first sample test is folowing:

• 1-substring: bcabcabcabcabca
• 2-substring: cabcabcabcabc
• 3-substring: abcabcabcab
• 4-substring: bcabcabca
• 5-substring: cabcabc
• 6-substring: abcab
• 7-substring: bca
• 8-substring: c

思路一：考虑用hash值判断字符串等价性。对于每个查询点，我们正常二分长度是不满足单调性的；于是我们转换思路，不对单点二分，而是对于每个固定的答案中心点，因为其对称的另外的点是固定的，因此可以二分求出最大匹配长度，如对于mi，最大匹配长度是2 * x + 1，则更新ans[mi - x] = max(ans[mi - x], 2 * x + 1)。这样枚举完中间点之后，我们并没有做完，因为我们只更新了每个中心点最大匹配长度对应的起点的答案，但相同的中心点，也可能作为其他起点取到最大值的中心点，因此对于每个ans[i]，先将其赋为max(ans[i], ans[i - 1] - 2)。这样扫一遍ans数组就行了。瓶颈在于预处理，nlogn。

#include <iostream>
#include <fstream>
#include <sstream>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <queue>
#include <stack>
#include <vector>
#include <set>
#include <map>
#include <list>
#include <iomanip>
#include <cctype>
#include <cassert>
#include <bitset>
#include <ctime>

using namespace std;

#define pau system("pause")
#define ll long long
#define pii pair<int, int>
#define pb push_back
#define mp make_pair
#define clr(a, x) memset(a, x, sizeof(a))

const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double EPS = 1e-9;
const int B = 131;

int n;
char s[1000015];
ll Hash[1000015], pp[1000015];
ll get_Hash(int s, int e) {
    ll res = (Hash[e] - pp[e - s + 1] * Hash[s - 1]) % MOD;
    if (res < 0) res += MOD;
    return res;
}
int ans[1000015];
bool check(int mi, int x) {
    return get_Hash(mi - x + 1, mi + x - 1) == get_Hash(n + 1 - mi - x + 1, n + 1 - mi + x - 1);
}
int main() {
    scanf("%d %s", &n, s + 1);
    pp[0] = 1;
    for (int i = 1; i <= n; ++i) {
        Hash[i] = (Hash[i - 1] * B + s[i]) % MOD;
        pp[i] = pp[i - 1] * B % MOD;
    }
    clr(ans, -1);
    for (int i = 1; i <= n + 1 >> 1; ++i) {
        int s = 1, e = i, mi, res = 0;
        if (2 * i == n + 1) break;
        while (s <= e) {
            mi = s + e >> 1;
            if (check(i, mi)) s = (res = mi) + 1;
            else e = mi - 1;
        }
        ans[i - res + 1] = max(ans[i - res + 1], 2 * res - 1);
    }
    for (int i = 1; i <= n + 1 >> 1; ++i) {
        ans[i] = max(ans[i - 1] - 2, ans[i]);
        printf("%d ", ans[i]);
    }
    return 0;
}

思路二：我们注意到性质ans[i] >= ans[i - 1] + 2, 也就是ans[i - 1] <= ans[i] - 2.因此我们从后往前扫ans数组，每个点从ans[i] - 2暴力向下枚举，找到答案就是最终解，并马上break。这样总体的步幅顶多n + 2 * n。复杂度O(n)。

#include <iostream>
#include <fstream>
#include <sstream>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <queue>
#include <stack>
#include <vector>
#include <set>
#include <map>
#include <list>
#include <iomanip>
#include <cctype>
#include <cassert>
#include <bitset>
#include <ctime>

using namespace std;

#define pau system("pause")
#define ll long long
#define pii pair<int, int>
#define pb push_back
#define mp make_pair
#define clr(a, x) memset(a, x, sizeof(a))

const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double EPS = 1e-9;
const int B = 131;

int n;
char s[1000015];
ll Hash[1000015], pp[1000015];
ll get_Hash(int s, int e) {
    ll res = (Hash[e] - pp[e - s + 1] * Hash[s - 1]) % MOD;
    if (res < 0) res += MOD;
    return res;
}
int ans[1000015];
bool check(int mi, int x) {
    return get_Hash(mi - x + 1, mi + x - 1) == get_Hash(n + 1 - mi - x + 1, n + 1 - mi + x - 1);
}
int main() {
    scanf("%d %s", &n, s + 1);
    pp[0] = 1;
    for (int i = 1; i <= n; ++i) {
        Hash[i] = (Hash[i - 1] * B + s[i]) % MOD;
        pp[i] = pp[i - 1] * B % MOD;
    }
    clr(ans, -1);
    for (int i = n + 1 >> 1; i; --i) {
        int l = i, r = n + 1 - i;
        if (l == r) {
            ans[i] = -1;
            continue;
        }
        for (int j = ans[i + 1] + 2; ~j; j -= 2) {
            if (get_Hash(l, l + j - 1) == get_Hash(r - j + 1, r)) {
                ans[i] = j;
                break;
            }
        }
    }
    for (int i = 1; i <= n + 1 >> 1; ++i) {
        printf("%d ", ans[i]);
    }
    return 0;
}