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 
    
     # 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\n", ans);    return 0;}