给一个串，让你找一个子串，形如EAEBE，就是一个串在开头结尾中间各出现一次，问这个E最长是多少
Sample Input
5
xy
abc
aaa
aaaaba
aaxoaaaaa
Sample Output
0
0
1
1
2
1 # include <cstdio> 2 # include <cstring> 3 using namespace std; 4 5 char S[1000010] ; 6 int next[1000010] ; 7 int slen ; 8 int tlen ; 9 10 11 void getNext() 12 { 13 int j, k; 14 j = 0; k = -1; next[0] = -1; 15 while(j < slen) 16 if(k == -1 || S[j] == S[k]) 17 next[++j] = ++k; 18 else 19 k = next[k]; 20 21 } 22 23 24 25 int main () 26 { 27 int T ; 28 scanf("%d" , &T); 29 while (T--) 30 { 31 scanf("%s" , S) ; 32 slen = strlen(S) ; 33 getNext() ; 34 int t = slen ; 35 int ans = 0 ; 36 while (t > slen/3) 37 t = next[t] ; 38 while(t) 39 { 40 tlen = t ; 41 int i ; 42 int flag = 1 ; 43 44 for(i = t*2 ; i <= slen - t ; i++)//利用next数组的性质 45 { 46 if (next[i] == tlen) 47 { 48 ans = tlen ; 49 flag = 0 ; 50 break ; 51 } 52 } 53 if (flag) 54 t = next[t] ; 55 else 56 break ; 57 } 58 printf("%d " , ans) ; 59 60 } 61 62 63 return 0 ; 64 } 65 66 67 68 69 # include <cstdio> 70 # include <cstring> 71 using namespace std; 72 73 char S[1000010] ; 74 int next[1000010] ; 75 int slen ; 76 int tlen ; 77 78 79 void getNext() 80 { 81 int j, k; 82 j = 0; k = -1; next[0] = -1; 83 while(j < slen) 84 if(k == -1 || S[j] == S[k]) 85 next[++j] = ++k; 86 else 87 k = next[k]; 88 89 } 90 91 bool kmp(int a , int b) 92 { 93 94 int i, j = 0; 95 96 97 for(i = a; i <= b; i++) 98 { 99 while(j > 0 && S[i] != S[j]) 100 j = next[j]; 101 if(S[i] == S[j]) 102 j++; 103 if(j == tlen) 104 { 105 return true ; 106 } 107 } 108 return false; 109 } 110 111 int main () 112 { 113 int T ; 114 scanf("%d" , &T); 115 while (T--) 116 { 117 scanf("%s" , S) ; 118 slen = strlen(S) ; 119 getNext() ; 120 int t = slen ; 121 int ans = 0 ; 122 while (t > slen/3) 123 t = next[t] ; 124 while(t) 125 { 126 tlen = t ; 127 int flag = kmp(t ,slen - t - 1) ;//查看模式串是否匹配主串 128 if (flag) 129 { 130 ans = t ; 131 break ; 132 } 133 t = next[t] ; 134 } 135 printf("%d " , ans) ; 136 137 } 138 139 140 return 0 ; 141 }