Period
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 1543 Accepted Submission(s): 746
Problem Description
For each prefix of a given string S with N characters (each character has an ASCII code between 97 and 126, inclusive), we want to know whether the prefix is a periodic string. That is, for each i (2 <= i <= N) we want to know the largest K > 1 (if there is one) such that the prefix of S with length i can be written as A K , that is A concatenated K times, for some string A. Of course, we also want to know the period K.
Input
The input file consists of several test cases. Each test case consists of two lines. The first one contains N (2 <= N <= 1 000 000) – the size of the string S. The second line contains the string S. The input file ends with a line, having the number zero on it.
Output
For each test case, output “Test case #” and the consecutive test case number on a single line; then, for each prefix with length i that has a period K > 1, output the prefix size i and the period K separated by a single space; the prefix sizes must be in increasing order. Print a blank line after each test case.
Sample Input
3aaa12aabaabaabaab0
Sample Output
Test case #12 23 3Test case #22 26 29 312 4
Recommend
思路：本题有用到KMP里面求next函数的一个结论，那就是当前的标号减去next的记录值就等于最小循环节的长度。
所以这题解就用next函数就可以处理了。
#include <iostream>#include <cstring>#include <map>using namespace std;const int mm=1e6+10;char s[mm];int next[mm],len;void get(){ memset(next,0,sizeof(next)); next[0]=-1;int k=-1; for(int i=0;i<len;) { if(k==-1||s[i]==s[k]) { ++k;++i; next[i]=k; }else k=next[k]; }}int main(){ int cas=0; while(cin>>len&&len) { cin>>s; get(); ///for(int i=0;i<=len;i++) ///cout<<next[i]<<" "; int temp; cout<<"Test case #"<<++cas<<"\n"; for(int i=2;i<=len;i++) { temp=i-next[i];///串循环长度 int ans=i/temp;///循环次数 if(i%temp==0&&ans>1) cout<<i<<" "<<ans<<"\n"; } cout<<"\n"; }}