很有意思的题，做完感觉脑子里一根筋又被拨回正轨了。

考虑一个做本质不同子串的通用思路。将该串的所有后缀排序，并统计\(\sum{n - sa_i + height_i}\)即可。

这样做本质上是容斥。但这个容斥与别人不同，他通过字典序来贪心地保证了答案正确，而不像随便一个顺序，那样的话不会容斥完全。

那么本题就有了些许做法了。

考虑先reverse一下这个串，这样的话后缀数组仅仅是增加一个量，也就有可能被我们维护。

进一步地去想，如果我们离线地先将后缀数组build出来，那么每次在这个串里面添加一个字符，就等价于添加一个后缀，就等价于往已经排好序的后缀“空”里面“填”。

这样就很简单了，搞个set来维护所有后缀排序后的下标，每次找前驱后继更新一下lcp之和即可。

注意两个后缀的lcp就是\(i=rk_i,j=rk_j\)后的区间height的min，这个可以st表做到\(\mathcal{O(1)}\).

#include <bits/stdc++.h>

#define test(...) fprintf(stderr, __VA_ARGS__)
#define dbg(x) cerr << #x << " = " << x << '\n'

using namespace std;

typedef long long ll;
typedef pair <int, int> pii;
typedef vector <int> vi;
typedef unsigned int ui;
typedef vector <pair <int, int> > edges; 

const int N = 100010 + 5; 
int sa[N], rk[N], rk_[N], id[N], px[N], cnt[N], h[N], n, a[N], b[N], SZ, LG[N];
int rmq[N][20];
struct node {
    int x;
    node(int x=0):x(x){}
    bool operator < (const node& rhs) const {
        return rk[x] < rk[rhs.x];
    }
};
bool cmp(int x, int y, int w) {
    return rk_[x] == rk_[y] && rk_[x + w] == rk_[y + w];
}
void re_order() {
    for (int i = 1; i <= n; ++i) b[i] = a[i];
    sort (b + 1, b + n + 1);
    SZ = unique(b + 1, b + n + 1) - (b + 1);
    for (int i = 1; i <= n; ++i) a[i] = lower_bound(b + 1, b + SZ + 1, a[i]) - b; 
}
void get_suffix_array() {
    int i, j, m = SZ, p, w;
    for (i = 1; i <= n; ++i) ++cnt[rk[i] = a[i]];
    for (i = 1; i <= m; ++i) cnt[i] += cnt[i - 1];
    for (i = n; i >= 1; --i) sa[cnt[rk[i]]--] = i;
    
    for (w = 1; w < n; w <<= 1, m = p) {
        for (p = 0, i = n; i > n - w; --i) id[++p] = i;
        for (i = 1; i <= n; ++i) if (sa[i] > w) id[++p] = sa[i] - w;
        memset (cnt, 0, sizeof(cnt));
        for (i = 1; i <= n; ++i) ++cnt[px[i] = rk[id[i]]];
        for (i = 1; i <= m; ++i) cnt[i] += cnt[i - 1];
        for (i = n; i >= 1; --i) sa[cnt[px[i]]--] = id[i];
        memcpy(rk_, rk, sizeof(rk));
        for (p = 0, i = 1; i <= n; ++i)
            rk[sa[i]] = (cmp(sa[i - 1], sa[i], w) ? p : ++p);
    }
    for (i = 1; i <= n; ++i) rk[sa[i]] = i; 
    for (i = 1, j = 0; i <= n; ++i) {
        if (rk[i] == 1) continue;
        while (a[i + j] == a[sa[rk[i] - 1] + j]) ++j;
        h[rk[i]] = j;
        rmq[rk[i]][0] = j;
        if (j) --j;
    }
}
int lcp(int i, int j) {
    i = rk[i], j = rk[j];
    if (i > j) swap(i, j);
    ++i;
    int depth = LG[j - i];
    return min(rmq[i][depth], rmq[j - (1 << depth) + 1][depth]);
}
void solve() {
    scanf ("%d", &n);
    for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
    for (int i = 2; i <= n; ++i) LG[i] = LG[i >> 1] + 1; 
    reverse(a + 1, a + n + 1);
    re_order();
    get_suffix_array();
    for (int j = 1; j <= 20; ++j)
        for (int i = 1; i <= n; ++i) {
            if (i + (1 << j) - 1 > n) break;
            rmq[i][j] = min(rmq[i][j - 1], rmq[i + (1 << (j - 1))][j - 1]);
        }   
    rk[0] = 0;
    rk[n + 1] = n + 1;
    set <node> st; 
    st.insert(0); 
    st.insert(n + 1);
    ll sum_lcp = 0; 
    ll total_length = 0;
    for (int i = 1; i <= n; ++i) {
        int pos = n - i + 1; 
        set <node> :: iterator it = st.lower_bound(node(pos));
        int nxt = (*it).x, prf = (*(--it)).x;
        sum_lcp += lcp(prf, nxt);
        sum_lcp -= lcp(prf, pos);
        sum_lcp -= lcp(pos, nxt);
        total_length += i;
        st.insert(node(pos));
        printf ("%lld\n", total_length + sum_lcp);
    }
} 

int main() {
    int tests = 1;
    while (tests--) 
        solve();
  return 0;
}