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\ [ans = \ sum_ {1
<= i <j <= n} len (T_i) + len (T_j) -2 * lcp (T_i, T_j) \] observed this formula can be found in the front two \ (len \) is a constant, is actually playing back the same suffix each multiplied by its prefix and the length.
Suffix is the same as the two two prefix string corresponding to the parent tree points \ (the LCA \) , then the statistics tree directly on the line.
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int MAXN = 1000010;
struct SAM{
int ch[26];
int len, fa;
}sam[MAXN << 1];
int las = 1, cnt = 1, f[MAXN << 1];
struct Edge{
int next, to;
}e[MAXN << 1];
int head[MAXN << 1], num, n;
inline void Add(int from, int to){
e[++num].to = to; e[num].next = head[from]; head[from] = num;
}
inline void add(int c){
int p = las; int np = las = ++cnt;
sam[np].len = sam[p].len + 1; f[cnt] = 1;
for(; p && !sam[p].ch[c]; p = sam[p].fa) sam[p].ch[c] = np;
if(!p) sam[np].fa = 1;
else{
int q = sam[p].ch[c];
if(sam[q].len == sam[p].len + 1) sam[np].fa = q;
else{
int nq = ++cnt; sam[nq] = sam[q];
sam[nq].len = sam[p].len + 1;
sam[q].fa = sam[np].fa = nq;
for(; p && sam[p].ch[c] == q; p = sam[p].fa) sam[p].ch[c] = nq;
}
}
}
char a[MAXN];
long long ans;
void dfs(int u){
long long tmp = 0;
for(int i = head[u]; i; i = e[i].next){
dfs(e[i].to);
tmp += (long long)f[u] * f[e[i].to];
f[u] += f[e[i].to];
}
ans += (long long)tmp * 2 * sam[u].len;
}
int main(){
scanf("%s", a + 1);
n = strlen(a + 1);
for(int i = 1; i <= n; ++i)
add(a[i] - 'a');
for(int i = 2; i <= cnt; ++i)
Add(sam[i].fa, i);
dfs(1);
printf("%lld\n", (long long)n * (n - 1) * (n + 1) / 2 - ans);
return 0;
}