Seek the Name, Seek the Fame

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 14103		Accepted: 7016

DescriptionThe 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 innovativelittle 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 stringsof S? (He might thank you by giving your baby a name:)InputThe 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.OutputFor 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

该题的题意是这样的，给若干个字符串，判断该字符串前n个字符和后n个字符是否相同，相同并按按从小到大的顺序输出n。

如果还有点困惑可以看一下这个链接有大神做的解题报告点击打开链接

#include<stdio.h>#include<string.h>char a[401000];int p[401000];int c[400100];int len;void kmp(){int i,j;i=0;j=-1;p[0]=-1;while(i<len){if(j==-1||a[i]==a[j]){i++;j++;p[i]=j;}else j=p[j];}}int main(){while(scanf("%s",a)!=EOF){len=strlen(a);kmp();int i,j,k=0;for(i=len;p[i]!=-1;i=p[i])c[k++]=i;printf("%d ",c[k-1]);for(j=k-2;j>=0;j--)printf("%d ",c[j]);printf("\n");}}

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