【回文自动机】 BZOJ 2160 拉拉队排练

通道

题意：求所有回文按长度排序后，前k个奇数长度的子串乘积之和。

思路：len[i]代表长度为i的有多少个子串，避免超时，树状数组维护即可

代码：

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

typedef long long ll;

const int MAX_N = 1200005;
const int SIG = 26 ;

const int N = MAX_N;

struct Bit {
    ll bit[N];
    void clear() {
        memset(bit, 0, sizeof bit);
    }
    void add(ll i, ll v) {
        if (i == 0) bit[0] += v;
        else {
            for (;i < N; i += i & -i)
                bit[i] += v;
        }
    }
    ll query(int i) {
        if (i < 0) return 0;
        else {
            ll re = 0;
            for (;i > 0; i -= i & -i)
                re += bit[i];
            return re;
        }
    }
};

Bit B;
struct PTree {
    int nxt[MAX_N][SIG], fail[MAX_N],  num[MAX_N], S[MAX_N];
    int last, n, p;ll cnt[MAX_N], len[MAX_N];
    int newNode (int l) {
        memset(nxt[p], 0, sizeof nxt[p]);
        cnt[p] = num[p] = 0;
        len[p] = l;
        return p++;
    }
    void init() {
        p = last = n = 0;
        newNode(0), newNode(-1);
        S[n] = -1;
        fail[0] = 1;
    }
    int getFail(int x) {
        while (S[n - len[x] - 1] != S[n]) x = fail[x];
        return x;
    }
    void add(int c) {
        c -= 'a';
        S[++n] = c;
        int cur = getFail(last);
        if (!nxt[cur][c]) {
            int now = newNode(len[cur] + 2);
            fail[now] = nxt[getFail(fail[cur])][c];
            nxt[cur][c] = now;
            num[now] = num[fail[now]] + 1;
        }
        last = nxt[cur][c];
        ++cnt[last];
    }    
    void count() {
        for (int i = p - 1; i >= 0; --i) cnt[fail[i]] += cnt[i];
    }
};

const ll MOD = 19930726ll;

PTree A;
ll n, m, len[MAX_N];
char s[MAX_N];

ll Pow(ll x, ll n) {
    ll ret = 1;
    while (n > 0) {
        if (n & 1) ret = ret * x % MOD;
        x = x * x % MOD;
        n >>= 1;
    }
    return ret;
}

int main() {
    A.init();
    scanf("%lld%lld%s", &n, &m, s);
    for (int i = 0; s[i]; ++i) A.add(s[i]);
    ll ans = 1;
    B.clear();
    A.count();
    for (int i = 2; i < A.p; ++i) {
        B.add(A.len[i], A.cnt[i]);
    }
    len[1] = B.query(1);
    for (int i = 2; i <= n; ++i) len[i] = B.query(i) - B.query(i - 1);
    for (int i = n; m && i > 0; --i) {
        if (i % 2 == 0 || len[i] == 0) continue;
        if (len[i] > m) {
            ans = (ans * Pow(i, m)) % MOD;
            m = 0;
        } else {
            ans = (ans * Pow(i, len[i])) % MOD;
            m -= len[i];
        }
    }
    if (m > 0) ans = -1;
    printf("%lld\n", ans);
    return 0;
}