1010 - The Minimum Length

Time Limit: 1s Memory Limit: 128MB

Submissions: 2502 Solved: 925

DESCRIPTION

There is a string A. The length of A is less than 1,000,000. I rewrite it again and again. Then I got a new string: AAAAAA...... Now I cut it from two different position and get a new string B. Then, give you the string B, can you tell me the length of the shortest possible string A. For example, A="abcdefg". I got abcdefgabcdefgabcdefgabcdefg.... Then I cut the red part: efgabcdefgabcde as string B. From B, you should find out the shortest A.

INPUT

Multiply Test Cases. For each line there is a string B which contains only lowercase and uppercase charactors. The length of B is no more than 1,000,000.

OUTPUT

For each line, output an integer, as described above.

SAMPLE INPUT

bcabcab
efgabcdefgabcde

SAMPLE OUTPUT

3
7
思路：kmp+最小循环节

 1 #include <bits/stdc++.h>
 2 
 3 using namespace std;
 4 
 5 #define MP make_pair
 6 #define PB push_back
 7 typedef long long LL;
 8 typedef pair<int,int> PII;
 9 const double eps=1e-8;
10 const double pi=acos(-1.0);
11 const int K=1e6+7;
12 const int mod=1e9+7;
13 
14 int nt[K];
15 char sa[K],sb[K];
16 
17 void kmp_next(char *T,int *nt)
18 {
19     nt[0]=0;
20     for(int i=1,j=0,len=strlen(T);i<len;i++)
21     {
22         while(j&&T[j]!=T[i])j=nt[j-1];
23         if(T[j]==T[i])j++;
24         nt[i]=j;
25     }
26 }
27 int kmp(char *S,char *T,int *nt)
28 {
29     int ans=0;
30     kmp_next(T,nt);
31     int ls=strlen(S),lt=strlen(T);
32     for(int i=0,j=0;i<ls;i++)
33     {
34         while(j&&S[i]!=T[j])j=nt[j-1];
35         if(S[i]==T[j])j++;
36         if(j==lt)
37              ans++,j=0;
38     }
39     return ans;
40 }
41 int main(void)
42 {
43     int t;
44     while(scanf("%s",sa)==1)
45     {
46         kmp_next(sa,nt);
47         int len=strlen(sa);
48         int ans=len-nt[len-1];
49         printf("%d
",ans);
50     }
51     return 0;
52 }