Seek the Name, Seek the Fame
Time Limit : 4000/2000ms (Java/Other) Memory Limit : 131072/65536K (Java/Other)
Total Submission(s) : 43 Accepted Submission(s) : 24
Problem 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
PKU
题目分析：
题目意思是让你输出所有首尾相同子串的长度 例如abab cabab ababcabab 有首ab尾ab 首 abab尾abab 首ababcabab尾 ababcabab ababcababababcabab
i 0 1 2 3 4 5 6
a b a b a b
j -1 0 0 1 2 3 4
p[6]=4,p[5]=3,p[4]=2,p[3]=1,p[2]=0,p[1]=0,p[0]=-1.
到6时匹配失败跳转到p[6]=4位置说明4 符合要求 p[4]=2 说明2符合 所以 有 2 4 6
#include<cstdio>
#include<cstring>
const int M=400000+10;
int len;
char str[M];
int p[M],a[M];
void getp()
{
int i=0,j=-1;
p[i]=j;
while(i<len)
{
if(j==-1||str[i]==str[j])
{
i++;j++;
p[i]=j;
}
else j=p[j];
}
}
int main()
{
while(~scanf("%s",str))
{
len=strlen(str);
getp();
int j=0;
for(int i=len;p[i]!=-1;i=p[i])// 此处就是统计相同首尾的长度
a[j++]=i;
for(int i=j-1;i>=0;i--)
printf("%d ",a[i]);
printf("\n");
}
return 0;
}