设串为str, 串长为len。
对整个串求一遍next函数，从串结尾开始顺着next函数往前找<=len/3的最长串，假设串长为ans，由于next的性质，所以找到的串肯定满足E……E这种形式，然后就是在str[ans]-str[len-2*ans]中查找是不是包含E，找到就输出，找不到就沿着next向下寻找，缩短串长。
#include#include #include #include using namespace std;const int MAXN = 1000100;char str[MAXN];char tmp[MAXN];int nextval[MAXN];int len;void GetNextVal( char *s, int lenth ){ int i = 0, j = -1; nextval[0] = -1; while ( i < lenth ) { if ( j == -1 || s[i] == s[j] ) { ++i, ++j; if ( s[i] != s[j] ) nextval[i] = j; else nextval[i] = nextval[j]; } else j = nextval[j]; } return;}bool KMP( char *t, char *s, int lenth, int lenn ){ //GetNextVal( t, lenth ); int i = 0, j = 0; while ( j < lenn ) { if ( i == -1 || s[j] == t[i] ) { ++i, ++j; if ( i == lenth ) return true; } else i = nextval[i]; //if ( i == lenth ) return true; } return false;}int main(){ int T; scanf( "%d", &T ); while ( T-- ) { scanf( "%s", str ); len = strlen(str); GetNextVal( str, len ); int ans = nextval[len]; while ( ans > len/3 ) ans = nextval[ans]; while ( ans > 0 ) { if ( KMP( str, &str[ans], ans, len - ans - ans ) ) break; ans = nextval[ans]; } if ( ans < 0 ) printf("%d\n", len/3); else printf( "%d\n", ans ); } return 0;}