题意
给定一个长度为n的小写字母串。问你有多少对相交的回文子串（包含也算相交）
相交的回文子串个数 (mod 51123987)
Sol
求相交的回文子串不太好求
考虑用总数减去不相交的回文串个数
那么考虑求以一个点结尾的后缀回文串的贡献：
就是以它后面的点为开头的前缀回文串的个数
正反两遍回文树求一下就好了
# include <bits/stdc++.h>
# define IL inline
# define RG register
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
IL int Input(){
RG int x = 0, z = 1; RG char c = getchar();
for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
return x * z;
}
const int maxn(2e6 + 5);
const int mod(51123987);
int first[maxn], nxt[maxn], type[maxn], fa[maxn], deep[maxn], pre[maxn], len[maxn];
int tot, last, ans, n;
char s[maxn];
IL void Upd(RG int &x, RG int y){
x += y;
if(x >= mod) x -= mod;
}
IL int NewNode(){
first[++tot] = len[tot] = fa[tot] = deep[tot] = nxt[tot] = 0;
type[tot] = -1;
return tot;
}
IL void Init(){
first[1] = first[0] = nxt[1] = nxt[0] = 0, type[0] = type[1] = -1;
fa[0] = fa[1] = 1, tot = 1, last = 0, len[1] = -1;
}
IL int Son(RG int u, RG int c){
for(RG int v = first[u]; v; v = nxt[v])
if(type[v] == c) return v;
return 0;
}
IL void Link(RG int u, RG int v, RG int c){
nxt[v] = first[u], first[u] = v, type[v] = c;
}
IL void Extend(RG int pos, RG int c){
RG int p = last;
while(s[pos - len[p] - 1] != s[pos]) p = fa[p];
if(!Son(p, c)){
RG int np = NewNode(), q = fa[p];
while(s[pos - len[q] - 1] != s[pos]) q = fa[q];
len[np] = len[p] + 2, fa[np] = Son(q, c);
Link(p, np, c), deep[np] = deep[fa[np]] + 1;
}
last = Son(p, c);
}
int main(){
n = Input(), scanf(" %s", s + 1), Init();
for(RG int i = 1; i <= n; ++i){
Extend(i, s[i] - 'a');
pre[i] = deep[last], Upd(pre[i], pre[i - 1]);
Upd(ans, deep[last]);
}
ans = (1LL * ans * (ans - 1) >> 1) % mod;
reverse(s + 1, s + n + 1), Init();
for(RG int i = 1; i <= n; ++i){
Extend(i, s[i] - 'a');
Upd(ans, mod - 1LL * deep[last] * pre[n - i] % mod);
}
printf("%d
", ans);
return 0;
}