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

  1 #include <set>
  2 #include <map>
  3 #include <stack>
  4 #include <queue>
  5 #include <cmath>
  6 #include <ctime>
  7 #include <cstdio>
  8 #include <string>
  9 #include <vector>
 10 #include <cstring>
 11 #include <iostream>
 12 #include <algorithm>
 13 #include <unordered_map>
 14 
 15 #define  pi    acos(-1.0)
 16 #define  eps   1e-9
 17 #define  fi    first
 18 #define  se    second
 19 #define  rtl   rt<<1
 20 #define  rtr   rt<<1|1
 21 #define  bug                printf("******\n")
 22 #define  mem(a, b)          memset(a,b,sizeof(a))
 23 #define  name2str(x)        #x
 24 #define  fuck(x)            cout<<#x" = "<<x<<endl
 25 #define  sfi(a)             scanf("%d", &a)
 26 #define  sffi(a, b)         scanf("%d %d", &a, &b)
 27 #define  sfffi(a, b, c)     scanf("%d %d %d", &a, &b, &c)
 28 #define  sffffi(a, b, c, d) scanf("%d %d %d %d", &a, &b, &c, &d)
 29 #define  sfL(a)             scanf("%lld", &a)
 30 #define  sffL(a, b)         scanf("%lld %lld", &a, &b)
 31 #define  sfffL(a, b, c)     scanf("%lld %lld %lld", &a, &b, &c)
 32 #define  sffffL(a, b, c, d) scanf("%lld %lld %lld %lld", &a, &b, &c, &d)
 33 #define  sfs(a)             scanf("%s", a)
 34 #define  sffs(a, b)         scanf("%s %s", a, b)
 35 #define  sfffs(a, b, c)     scanf("%s %s %s", a, b, c)
 36 #define  sffffs(a, b, c, d) scanf("%s %s %s %s", a, b,c, d)
 37 #define  FIN                freopen("../in.txt","r",stdin)
 38 #define  gcd(a, b)          __gcd(a,b)
 39 #define  lowbit(x)          x&-x
 40 #define  IO                 iOS::sync_with_stdio(false)
 41 #pragma comment(linker, "/STACK:102400000,102400000")
 42 
 43 using namespace std;
 44 typedef long long LL;
 45 typedef unsigned long long ULL;
 46 const ULL seed = 13331;
 47 const LL INFLL = 0x3f3f3f3f3f3f3f3fLL;
 48 const int maxn = 1e6 + 3;
 49 const int maxm = 8e6 + 10;
 50 const int INF = 0x3f3f3f3f;
 51 const int mod = 1e9 + 7;
 52 char s[maxn];
 53 int sum1[maxn];
 54 
 55 struct Palindrome_Automaton {
 56     int len[maxn], next[maxn][26], fail[maxn], sum[maxn];
 57     int num[maxn], S[maxn], sz, n, last;
 58 
 59     int newnode(int l) {
 60         for (int i = 0; i < 26; ++i)next[sz][i] = 0;
 61         num[sz] = 0, len[sz] = l;
 62         return sz++;
 63     }
 64 
 65     void init() {
 66         sz = n = last = 0;
 67         newnode(0);
 68         newnode(-1);
 69         S[0] = -1;
 70         fail[0] = 1;
 71     }
 72 
 73     int get_fail(int x) {
 74         while (S[n - len[x] - 1] != S[n])x = fail[x];
 75         return x;
 76     }
 77 
 78     void add(int c) {
 79         c -= 'a';
 80         S[++n] = c;
 81         int cur = get_fail(last);
 82         if (!next[cur][c]) {
 83             int now = newnode(len[cur] + 2);
 84             fail[now] = next[get_fail(fail[cur])][c];
 85             next[cur][c] = now;
 86             num[now] = num[fail[now]] + 1;
 87             sum[now] = (sum[fail[now]] + len[now]) % mod;
 88         }
 89         last = next[cur][c];
 90     }
 91 
 92 } pam;
 93 
 94 int main() {
 95     //FIN;
 96     while (~sfs(s+1)) {
 97         int n = strlen(s + 1);
 98         pam.init();
 99         for (int i = 1; i<=n; ++i) {
100             pam.add(s[i]);
101             sum1[i] = (1LL * (i + 1) * pam.num[pam.last] % mod - pam.sum[pam.last] + mod) % mod;
102         }
103         LL ans = 0;
104         pam.init();
105         for (int i = n; i >= 1; i--) {
106             pam.add(s[i]);
107             ans = (ans + sum1[i - 1] * (1LL * (i - 1) * pam.num[pam.last] % mod + pam.sum[pam.last]) % mod) % mod;
108         }
109         printf("%lld\n", ans);
110     }
111     return 0;
112 }

View Code