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Seek the Name, Seek the Fame
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. 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 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
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题意：
给你一个字符串。要你找出既是前缀又是后缀的子串长度的可能值。按字典序输出。
思路：
转化下思路其实很简单。既然既是前缀又是后缀。直接用文本串构造一个失配数组。然后直接匹配文本串的文本末。即txt[n].n=strlen(n)。而长度就为f[j]。想一想就明白了。感觉思想还是蛮好的。
详细见代码：
- #include <iostream>
- #include<string.h>
- #include<stdio.h>
- using namespace std;
- char txt[400100];
- int f[400100],ans[400100],cnt;
- void getf(char *p)//得到失配数组
- {
- int i,j,m=strlen(p);
- f[0]=f[1]=0;
- for(i=1;i<m;i++)
- {
- j=f[i];
- while(j&&p[j]!=p[i])
- j=f[j];
- f[i+1]=p[j]==p[i]?j+1:0;
- }
- }
- int main()
- {
- int len,i,j;
- while(~scanf("%s",txt))
- {
- getf(txt);
- len=strlen(txt);
- cnt=0;
- j=len;
- while(f[j])
- {
- ans[cnt++]=f[j];
- j=f[j];
- }
- for(i=cnt-1;i>=0;i--)
- printf("%d ",ans[i]);
- printf("%d\n",len);//注意自己是自己的前后缀
- }
- return 0;
- }
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Power Strings
Description
Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b = "abcdef". If we think of concatenation as multiplication, exponentiation by a non-negative integer is defined in the normal way: a^0 = "" (the
empty string) and a^(n+1) = a*(a^n). Input
Each test case is a line of input representing s, a string of printable characters. The length of s will be at least 1 and will not exceed 1 million characters. A line containing a period follows the last test case.
Output
For each s you should print the largest n such that s = a^n for some string a.
Sample Input abcd aaaa ababab . Sample Output 1 4 3 Hint
This problem has huge input, use scanf instead of cin to avoid time limit exceed.
Source |
题意：
给你一个字符串。问你这个字符串最多可以由一个子串重复多少次得到。
思路：
和大白(刘汝佳 训练指南)类似。先获取失配数组。然后匹配文本尾。len-f[i]就为循环节长度。
详细见代码：
- #include <iostream>
- #include<stdio.h>
- #include<string.h>
- using namespace std;
- const int maxn=1000100;
- int f[maxn];
- char txt[maxn];
- void getf(char *p)
- {
- int i,j,m=strlen(p);
- f[0]=f[1]=0;
- for(i=1;i<m;i++)
- {
- j=f[i];
- while(j&&p[i]!=p[j])
- j=f[j];
- f[i+1]=p[i]==p[j]?j+1:0;
- }
- }
- int main()
- {
- int len,i,ans,t;
- while(~scanf("%s",txt))
- {
- if(txt[0]=='.')
- break;
- getf(txt);
- len=strlen(txt);
- i=len;
- ans=1;
- while(f[i])
- {
- t=len-f[i];
- if(len%t==0&&len/t>ans)
- ans=len/t;
- i=f[i];
- }
- printf("%d\n",ans);
- }
- return 0;
-
}