后缀数组，SA

主要是 \(O(n\log n)\) 倍增求 SA。

(为什么这么短)

const int N = 1e6 + 9;
int n;
int sa[N], sa_tmp[N], rk[N], ork[N], buc[N], ht[N];
char s[N];

void getSA() {
    int m = 127;
    rep (i, 1, n) ++buc[rk[i] = s[i]];
    rep (i, 1, m) buc[i] += buc[i - 1];
    per (i, n, 1) sa[buc[rk[i]]--] = i;

    for (int len = 1, p = 0; ; m = p, p = 0, len *= 2) {
        rep (i, n - len + 1, n) sa_tmp[++p] = i;
        // 这里不用判断吗
        rep (i, 1, n) if (sa[i] > len) sa_tmp[++p] = sa[i] - len;

        memset(buc, 0, sizeof(buc[0]) * (m + 1));
        rep (i, 1, n) ++buc[rk[i]];
        rep (i, 1, m) buc[i] += buc[i - 1];
        per (i, n, 1) {
            int x = sa_tmp[i];
            sa[buc[rk[x]]--] = x;
        }

        rep (i, 1, n) ork[i] = rk[i];
        p = 0;
        rep (i, 1, n) rk[sa[i]] = (ork[sa[i - 1]] == ork[sa[i]] && ork[sa[i - 1] + len] == ork[sa[i] + len]) ? p : ++p;
        if (p == n) break;
    }

    int j = 0;
    rep (i, 1, n - 1) {
        if (j) --j;
        while (s[i + j] == s[sa[rk[i] - 1] + j]) ++j;
        ht[rk[i]] = j;
    }
}