题意：
输出每个长度下的回文串(这些回文串的左半边也需要是回文串)
题解：
直接套上回文树，然后每找到一个回文串就将他hash。
因为符合要求的回文串本身是回文串，左半边也是回文串，所以它左半边也右半边相同，
自己画个图很容易发现的
每次hash判断一下就好了
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("****** ") 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 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 + 7; 49 const int maxm = 8e6 + 10; 50 const int INF = 0x3f3f3f3f; 51 const int mod = 1e9 + 7; 52 char s[maxn]; 53 ULL p[maxn], HA[maxn]; 54 int ans[maxn]; 55 56 ULL get_HASH(int l, int r) { 57 return HA[r] - HA[l - 1] * p[r - l + 1]; 58 } 59 60 struct Palindrome_Automaton { 61 int len[maxn], next[maxn][26], fail[maxn], cnt[maxn]; 62 int num[maxn], str[maxn], sz, n, last; 63 int vis[maxn]; 64 65 int newnode(int l) { 66 for (int i = 0; i < 26; ++i)next[sz][i] = 0; 67 cnt[sz] = num[sz] = 0, len[sz] = l; 68 return sz++; 69 } 70 71 void init() { 72 sz = n = last = 0; 73 newnode(0); 74 newnode(-1); 75 str[0] = -1; 76 fail[0] = 1; 77 mem(vis, 0); 78 } 79 80 int get_fail(int x) { 81 while (str[n - len[x] - 1] != str[n])x = fail[x]; 82 return x; 83 } 84 85 void add(int c, int pos) { 86 c -= 'a'; 87 str[++n] = c; 88 int cur = get_fail(last); 89 if (!next[cur][c]) { 90 int now = newnode(len[cur] + 2); 91 fail[now] = next[get_fail(fail[cur])][c]; 92 next[cur][c] = now; 93 num[now] = num[fail[now]] + 1; 94 int temp = (len[now] + 1) / 2; 95 // fuck(n - len[now] + 1),fuck(n - len[now] + temp),fuck(n - temp),fuck(n); 96 if (len[now] == 1 || 97 get_HASH(n - len[now] + 1, n - len[now] + temp) == get_HASH(n - temp+1, n)) 98 vis[now] = 1; 99 else vis[now] = 0; 100 } 101 last = next[cur][c]; 102 cnt[last]++; 103 } 104 105 void count()//统计本质相同的回文串的出现次数 106 { 107 for (int i = sz - 1; i >= 0; --i) { 108 cnt[fail[i]] += cnt[i]; 109 if (vis[i]) ans[len[i]] += cnt[i]; 110 } 111 //逆序累加，保证每个点都会比它的父亲节点先算完，于是父亲节点能加到所有子孙 112 } 113 } pam; 114 115 int main() { 116 // FIN; 117 p[0] = 1; 118 for (int i = 1; i < maxn; ++i) p[i] = p[i - 1] * seed; 119 while (~sfs(s + 1)) { 120 int n = strlen(s + 1); 121 pam.init(); 122 for (int i = 1; i <= n; ++i) { 123 HA[i] = HA[i - 1] * seed + s[i]; 124 pam.add(s[i], i); 125 ans[i] = 0; 126 } 127 pam.count(); 128 for (int i = 1; i <= n; ++i) printf("%d%c", ans[i], (i == n ? ' ' : ' ')); 129 } 130 return 0; 131 }