hust1010 The Minimum Length

地址：http://acm.hust.edu.cn/problem/show/1010

题目：

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

 #include <bits/stdc++.h>

 using namespace std;

 #define MP make_pair
 #define PB push_back
 typedef long long LL;
 typedef pair<int,int> PII;
 const double eps=1e-;
 const double pi=acos(-1.0);
 const int K=1e6+;
 const int mod=1e9+;

 int nt[K];
 char sa[K],sb[K];

 void kmp_next(char *T,int *nt)
 {
     nt[]=;
     for(int i=,j=,len=strlen(T);i<len;i++)
     {
         while(j&&T[j]!=T[i])j=nt[j-];
         if(T[j]==T[i])j++;
         nt[i]=j;
     }
 }
 int kmp(char *S,char *T,int *nt)
 {
     int ans=;
     kmp_next(T,nt);
     int ls=strlen(S),lt=strlen(T);
     for(int i=,j=;i<ls;i++)
     {
         while(j&&S[i]!=T[j])j=nt[j-];
         if(S[i]==T[j])j++;
         if(j==lt)
              ans++,j=;
     }
     return ans;
 }
 int main(void)
 {
     int t;
     while(scanf("%s",sa)==)
     {
         kmp_next(sa,nt);
         int len=strlen(sa);
         int ans=len-nt[len-];
         printf("%d\n",ans);
     }
     return ;
 }