poj1961之KMP应用

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Period

Time Limit: 3000MS Memory Limit: 30000KTotal Submissions: 10998 Accepted: 5063

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 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

题目意思是给定一个长为n的字符串求字符串的前几位是循环字串并且求出循环次数

思路：对于一个长度为len的字符串,其可能周期=len-(len-next[len]),这个周期其实是多次循环后包含原串,并不是等于原串,所以要len%(len-next[len]) == 0来判断是否等于原串

这个所谓的周期是多次循环后包含原串,例如aaaabbaa=>求得周期是aaaabb

#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<string>#include<queue>#include<algorithm>#include<map>#include<iomanip>#define INF 99999999using namespace std;const int MAX=1000000+10;char s[MAX];int next[MAX];void get_next(char *a){int i=-1,j=0,len=strlen(a);next[0]=-1;while(j<len){if(i == -1 || a[i] == a[j])next[++j]=++i;else i=next[i];}for(int i=2;i<=len;++i){//前i个字符,i-next[i]是前i个字符的周期 if(i%(i-next[i]) == 0 && i/(i-next[i])>1)cout<<i<<' '<<i/(i-next[i])<<endl;}return;}int main(){int n,num=0;while(cin>>n,n){cin>>s;cout<<"Test case #"<<++num<<endl;get_next(s);cout<<endl;}return 0;}