码迷,mamicode.com
首页 > 其他好文 > 详细

FZU 1901 Period II

时间:2014-07-27 22:27:09      阅读:249      评论:0      收藏:0      [点我收藏+]

标签:des   style   blog   color   os   io   for   art   

Problem Description

For each prefix with length P of a given string S,if

S[i]=S[i+P] for i in [0..SIZE(S)-p-1],

then the prefix is a “period” of S. We want to all the periodic prefixs.

 

Input

Input contains multiple cases.

The first line contains an integer T representing the number of cases. Then following T cases.

Each test case contains a string S (1 <= SIZE(S) <= 1000000),represents the title.S consists of lowercase ,uppercase letter.

 

Output

For each test case, first output one line containing "Case #x: y", where x is the case number (starting from 1) and y is the number of periodic prefixs.Then output the lengths of the periodic prefixs in ascending order.
 
Sample Input
4 ooo acmacmacmacmacma fzufzufzuf stostootssto
 
Sample Output
Case #1: 3 1 2 3 Case #2: 6 3 6 9 12 15 16 Case #3: 4 3 6 9 10 Case #4: 2 9 12

 

•题意:给你一个字符串str,对于每个str长度为p的前缀,如果str[i]==str[p+i](p+i<len),那么我们认为它是一个periodic prefixs.求所有满足题意的前缀的长度p。
•知识点:KMP算法、对next数组的理解
•KMP算法中next数组的含义是什么?
•next数组:失配指针
•如果目标串的当前字符i在匹配到模式串的第j个字符时失配,那么我们可以让i试着去匹配next(j)
•对于模式串str,next数组的意义就是:
•如果next(j)=t,那么str[1…t]=str[len-t+1…len]
•我们考虑next(len),令t=next(len);
•next(len)有什么含义?
•str[1…t]=str[len-t+1…len]
•那么,长度为len-next(len)的前缀显然是符合题意的。
•接下来我们应该去考虑谁?
•t=next( next(len) );
•t=next( next (next(len) ) );
• 一直下去直到t=0,每个符合题意的前缀长是len-t
 
 1 #include <cstdio>
 2 #include <cstring>
 3 const int maxn = 1000010;
 4 int p[maxn],ans[maxn];
 5 char str[maxn];
 6 
 7 void get_p(int len){
 8     p[1] = 0;
 9     int j = 0;
10     for(int i = 2;i <= len;i++){
11         while(j > 0 && str[j+1] != str[i])
12             j = p[j];
13         if(str[j+1] == str[i]) 
14             j++;
15         p[i] = j;
16     }
17 }
18 
19 int main(){
20     int nkase;
21     scanf("%d",&nkase);
22     for(int kase = 1;kase <= nkase;kase++){
23         scanf("%s",str+1);
24         int len = strlen(str+1);
25         get_p(len);
26         int t = p[len],cnt = 0;
27         while(t){
28             ans[cnt++] = len-t;
29             t = p[t];
30         }
31         ans[cnt++] = len;
32         printf("Case #%d: %d\n",kase,cnt);
33         for(int i = 0;i < cnt-1;i++) 
34             printf("%d ",ans[i]);
35         printf("%d\n",ans[cnt-1]);
36     }
37     return 0;
38 }                    

 

FZU 1901 Period II,布布扣,bubuko.com

FZU 1901 Period II

标签:des   style   blog   color   os   io   for   art   

原文地址:http://www.cnblogs.com/zzy9669/p/3871446.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!