POJ 2752 Seek the Name, Seek the Fame kmp mismatch function next application

POJ 2752 Seek the Name, Seek the Fame kmp mismatch function next application
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Seek the Name, Seek the Fame Time Limit:2000 MS Memory Limit:65536 K Total Submissions:12791 Accepted:6304 Description The little cat is so famous, that could 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, cocould 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 ababcababababcababaaaaa Sample Output 2 4 9 181 2 3 4 5 Source POJ Monthly -- 2006.01.22, Zeyuan Zhu

Give you a string and find all the cases where the prefix is equal to the suffix. In the next Array application, KMP finds the next array. Each time the next [I] To I is removed, the remaining strings still meet the conditions until the header is found. Next [I]! If it is 0, it is the number of characters with the same prefix and suffix of the mode string.

//3892K547MS#include
 
  #include
  
   #define M 10007char pattern[400007];int next[400007],m,ans[400007];void pre(int len){    int i = 0, j = -1;    next[0] = -1;    while(i != len)    {        if(j == -1 || pattern[i] == pattern[j])            next[++i] = ++j;        else            j = next[j];    }}int main(){    while(scanf("%s",pattern)!=EOF)    {        memset(ans,0,sizeof(ans));        memset(next,0,sizeof(next));        m=strlen(pattern);        pre(m);        int k=0;        for(int i=m; i!=0; )        {            ans[k++]=next[i];            i=next[i];        }        for(int i=k-2; i>=0; i--)            printf("%d ",ans[i]);        printf("%d\n",m);    }    return 0;}