Seek the Name, Seek the Fame（KMP基础）

最新推荐文章于 2022-08-23 22:16:41 发布

原创最新推荐文章于 2022-08-23 22:16:41 发布 · 273 阅读

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本文介绍了一种用于为新生宝宝起名的算法，该算法利用父母的名字生成新的字符串，并从中寻找合适的前缀-后缀子串作为宝宝的名字。文章提供了一个具体的实现示例，包括输入输出格式及样例。

D - Seek the Name, Seek the Fame

Time Limit: 2000 MS Memory Limit: 65536 KB

64-bit integer IO format: %I64d , %I64u Java class name: Main

[Submit] [Status]

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

#include<stdio.h>
#include<string.h>
const int maxn = 400010;
char s[maxn];
int d[maxn];
int next[maxn];
int m;
void getnext()
{
    int i = 0, j = next[0] = -1;
    while(i < m)
    {
        if (j == -1 || s[i] == s[j])
        {
            ++i;
            ++j;
            next[i] = j;
        }
        else
            j = next[j];
    }
}

int main()
{
    while(~scanf("%s",s))
    {
        m = strlen(s);
        getnext();
        int c = 0;
        for(int i=m; next[i]!=-1; i=next[i])
        {
            d[c++] = i;
        }
        for(int i=c-1; i>=0; i--)
        {
            if(i == 0)
            {
                printf("%d\n",d[i]);
            }
            else
            {
                printf("%d ",d[i]);
            }
        }
    }
    return 0;
}