题目大意:给出一个字符串,求其不包含两个长度为k的"重复串"的最长前缀.
对于任意串s,有
(1)大写字母与小写字母等价.
(2)串中所有字母个数定义为串的长度.
两个串s1,s2为长度为k的重复串,定义为:
(1)s1,s2长度均为k
(2)对于每个字母,在s1中出现次数等于在s2中出现次数.
hash一下随便搞搞就好了,字符串hash记录一下值对应的字符串就好了
千万别用(439,479,607,739,487,751,433,739,317,509,541,281,337,199,787,23,859,181,281,373,673,701,421,373,457,283)
来加密……随机100~maxn之间的数就好了
贡献了无数tle的人的忠告
program poj2119; const maxn=999997; var n,ans,i,j,k,o:longint; loop:char; s,t:ansistring; sp:array ['a'..'z'] of longint; con:array ['a'..'z'] of longint; map:array [0..100001] of longint; rehash:array [0..maxn] of string[50]; count:array [0..maxn] of longint; function hash (o:longint):longint;inline; var i,k:longint; loop:char; now:string; begin k:=0; now:=''; for loop:='a' to 'z' do begin k:=(k+con[loop]*sp[loop]) mod maxn; for j:=1 to con[loop] do now:=now+loop; end; while (count[k]<>0)and(rehash[k]<>now) do begin inc(k); if k=maxn then k:=0; end; rehash[k]:=now; exit(k); end; begin randomize; for loop:='a' to 'z' do sp[loop]:=random(maxn-100)+90; repeat readln(k); if k=0 then break; fillchar(count,sizeof(count),0); fillchar(con,sizeof(con),0); readln(s); ans:=length(s); t:=''; n:=0; for i:=1 to ans do if (s[i]>='A')and(s[i]<='Z') then begin t:=t+chr(ord(s[i])+32); inc(n); map[n]:=i; end else if (s[i]>='a')and(s[i]<='z') then begin t:=t+s[i]; inc(n); map[n]:=i; end; if n<k then begin writeln(ans); continue; end; for i:=1 to k-1 do inc(con[t[i]]); for i:=k to n do begin inc(con[t[i]]); if i>k then dec(con[t[i-k]]); j:=hash(i); inc(count[j]); if count[j]=2 then begin ans:=map[i]-1; break; end; end; writeln(ans); until false; end.