Interesting HDU - 5785 回文树

题意：

找出所有【i，j】为回文串【j+1，k】也为回文串的i*k乘积之和。

题解：

设sum1【i】 为正着插入，到 i 的所有回文串的起始位置的前缀和，sum2【i】 表示反正插入的前缀和

ans+=sum1【i]*sum1【i+1】

上面的式子很容易让我们想到两遍回文树正着和反着插入操作，

回文树的num【】表示到达 i 这个节点的回文串个数

我们用一个sum【i】数组到 i 的时候所有出现的回文串的长度的前缀和

sum1[i] = (1LL * (i + 1) * pam.num[pam.last] % mod - pam.sum[pam.last] + mod) % mod;

由于卡内存 ，所以我们就不记录sum2[]了

 #include <set>
 #include <map>
 #include <stack>
 #include <queue>
 #include <cmath>
 #include <ctime>
 #include <cstdio>
 #include <string>
 #include <vector>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 #include <unordered_map>

 #define  pi    acos(-1.0)
 #define  eps   1e-9
 #define  fi    first
 #define  se    second
 #define  rtl   rt<<1
 #define  rtr   rt<<1|1
 #define  bug                printf("******\n")
 #define  mem(a, b)          memset(a,b,sizeof(a))
 #define  name2str(x)        #x
 #define  fuck(x)            cout<<#x" = "<<x<<endl
 #define  sfi(a)             scanf("%d", &a)
 #define  sffi(a, b)         scanf("%d %d", &a, &b)
 #define  sfffi(a, b, c)     scanf("%d %d %d", &a, &b, &c)
 #define  sffffi(a, b, c, d) scanf("%d %d %d %d", &a, &b, &c, &d)
 #define  sfL(a)             scanf("%lld", &a)
 #define  sffL(a, b)         scanf("%lld %lld", &a, &b)
 #define  sfffL(a, b, c)     scanf("%lld %lld %lld", &a, &b, &c)
 #define  sffffL(a, b, c, d) scanf("%lld %lld %lld %lld", &a, &b, &c, &d)
 #define  sfs(a)             scanf("%s", a)
 #define  sffs(a, b)         scanf("%s %s", a, b)
 #define  sfffs(a, b, c)     scanf("%s %s %s", a, b, c)
 #define  sffffs(a, b, c, d) scanf("%s %s %s %s", a, b,c, d)
 #define  FIN                freopen("../in.txt","r",stdin)
 #define  gcd(a, b)          __gcd(a,b)
 #define  lowbit(x)          x&-x
 #define  IO                 iOS::sync_with_stdio(false)
 #pragma comment(linker, "/STACK:102400000,102400000")

 using namespace std;
 typedef long long LL;
 typedef unsigned long long ULL;
 const ULL seed = ;
 const LL INFLL = 0x3f3f3f3f3f3f3f3fLL;
 const int maxn = 1e6 + ;
 const int maxm = 8e6 + ;
 const int INF = 0x3f3f3f3f;
 const int mod = 1e9 + ;
 char s[maxn];
 int sum1[maxn];

 struct Palindrome_Automaton {
     int len[maxn], next[maxn][], fail[maxn], sum[maxn];
     int num[maxn], S[maxn], sz, n, last;

     int newnode(int l) {
         for (int i = ; i < ; ++i)next[sz][i] = ;
         num[sz] = , len[sz] = l;
         return sz++;
     }

     void init() {
         sz = n = last = ;
         newnode();
         newnode(-);
         S[] = -;
         fail[] = ;
     }

     int get_fail(int x) {
         while (S[n - len[x] - ] != S[n])x = fail[x];
         return x;
     }

     void add(int c) {
         c -= 'a';
         S[++n] = c;
         int cur = get_fail(last);
         if (!next[cur][c]) {
             int now = newnode(len[cur] + );
             fail[now] = next[get_fail(fail[cur])][c];
             next[cur][c] = now;
             num[now] = num[fail[now]] + ;
             sum[now] = (sum[fail[now]] + len[now]) % mod;
         }
         last = next[cur][c];
     }

 } pam;

 int main() {
     //FIN;
     while (~sfs(s+)) {
         int n = strlen(s + );
         pam.init();
         for (int i = ; i<=n; ++i) {
             pam.add(s[i]);
             sum1[i] = (1LL * (i + ) * pam.num[pam.last] % mod - pam.sum[pam.last] + mod) % mod;
         }
         LL ans = ;
         pam.init();
         for (int i = n; i >= ; i--) {
             pam.add(s[i]);
             ans = (ans + sum1[i - ] * (1LL * (i - ) * pam.num[pam.last] % mod + pam.sum[pam.last]) % mod) % mod;
         }
         printf("%lld\n", ans);
     }
     return ;
 }