Seek the Name, Seek the Fame
| Time Limit: 2000MS | Memory Limit: 65536K | |
Description
The little cat is so famous, that many couples tramp over hill and dale to Byteland, and asked the little cat to give names to their newly-born babies. They seek the name, and at the same time seek the fame. In order to escape from such boring job, the innovative little cat works out an easy but fantastic algorithm:
Step1. Connect the father‘s name and the mother‘s name, to a new string S.
Step2. Find a proper prefix-suffix string of S (which is not only the prefix, but also the suffix of S).
Example: Father=‘ala‘, Mother=‘la‘, we have S = ‘ala‘+‘la‘ =
‘alala‘. Potential prefix-suffix strings of S are {‘a‘, ‘ala‘, ‘alala‘}.
Given the string S, could you help the little cat to write a program to
calculate the length of possible prefix-suffix strings of S? (He might
thank you by giving your baby a name:)
Input
The input contains a number of test cases. Each test case occupies a single line that contains the string S described above.
Restrictions: Only lowercase letters may appear in the input. 1 <= Length of S <= 400000.
Output
For
each test case, output a single line with integer numbers in increasing
order, denoting the possible length of the new baby‘s name.
Sample Input
ababcababababcabab aaaaa
Sample Output
2 4 9 18 1 2 3 4 5
Source
POJ Monthly--2006.01.22,Zeyuan Zhu
根据next函数的性质直接倒推就行了。。
1 #include<set>
2 #include<queue>
3 #include<cstdio>
4 #include<cstdlib>
5 #include<cstring>
6 #include<iostream>
7 #include<algorithm>
8 using namespace std;
9 const int N = 400010;
10 #define For(i,n) for(int i=1;i<=n;i++)
11 #define Rep(i,l,r) for(int i=l;i<=r;i++)
12 #define Down(i,r,l) for(int i=r;i>=l;i--)
13
14 char s[N];
15 int next[N],n,ans[N];
16
17 void BuildNext(char s[]){
18 next[0]=next[1]=0;
19 For(i,n-1){
20 int j=next[i];
21 while(j&&s[i]!=s[j]) j=next[j];
22 if(s[i]==s[j]) next[i+1]=j+1;
23 else next[i+1]=0;
24 }
25 }
26
27 int main(){
28 while(scanf("%s",&s)!=EOF){
29 n=strlen(s);BuildNext(s);
30 int p=next[n];ans[ans[0]=1,1]=n;
31 while(p){
32 ans[++ans[0]]=p;
33 p=next[p];
34 }
35 Down(i,ans[0],2) printf("%d ",ans[i]);printf("%d\n",ans[1]);
36 }
37 return 0;
38 }