P7114 字符串匹配

发表于 | 阅读数

求 $S = {(AB)}^iC$的方案数，其中 $F(A) \le F(C)$ 表示字符串 $S$ 中出现奇数次的字符的数量

枚举$(AB)^i$，对于$C$寻找符合条件的$AB$.

复杂度为$O(n(26+\log n))$

代码 
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#include <bits/stdc++.h>
using namespace std;

const int N = 2e6 + 10;

#define ll long long
void getnext(char *t, int *nxt)
{
    nxt[0] = -1;
    int m = strlen(t + 1);
    for (int i = 1, j; i <= m; i++)
    {
        for (j = nxt[i - 1]; j >= 0 && t[j + 1] != t[i]; j = nxt[j])
            ;
        nxt[i] = j + 1;
    }
}

char t[N], s[N];
int nxt[N];
int pre[N], suf[N], cnt[27], len[N];
int to[27];
int main()
{
    int T;
    scanf("%d", &T);
    while (T--)
    {
        scanf("%s", s + 1);
        int n = strlen(s + 1);

        for (int i = 1; i <= n; i++)
            nxt[i] = 0, len[i] = 1;
        getnext(s, nxt);
        memset(cnt, 0, sizeof(cnt));
        memset(to, 0, sizeof(to));
        int res = 0;
        for (int i = 1; i <= n; i++)
        {
            int o = s[i] - 'a';
            res -= cnt[o] & 1;
            cnt[o]++;
            res += cnt[o] & 1;
            pre[i] = res;
        }

        memset(cnt, 0, sizeof(cnt));
        res = 0;
        for (int i = n; i >= 1; i--)
        {
            int o = s[i] - 'a';
            res -= cnt[o] & 1;
            cnt[o]++;
            res += cnt[o] & 1;
            suf[i] = res;
        }

        ll ans = 0;
        for (int i = 2; i < n; i++)
        {

            for (int j = pre[i - 1]; j <= 26; j++)
                to[j]++;
            ans += to[suf[i + 1]];
            for (int j = 2 * i; j < n; j += i)
            {
                if (((i % (j - nxt[j])) == 0) && ((j / (j - nxt[j])) > 1))
                    ans += to[suf[j + 1]];
                else
                    break;
            }
        }
        printf("%lld\n", ans);
    }
}