poj 1961 Period

题目链接：http://poj.org/problem?id=1961

Period

Time Limit: 3000MS Memory Limit: 30000KTotal Submissions: 13197 Accepted: 6197

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

Source

Southeastern Europe 2004

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求到第i位的循环节，poj2406在最外边又加了一个循环而已。

#include<iostream>#include<cstring>#include<cmath>#include<cstdio>#include<algorithm>#include<cstdlib>#include<map>#include<set>#include<vector>#include<string>#include<stack>#include<queue>#include<bitset>using namespace std;#define CLR  memset(A,0,sizeof(A))const int MAX=1000005;int fail[MAX];char str[MAX];void get_next(char *pat){    memset(fail,-1,sizeof(fail));    for(int i=1;pat[i];++i){        int k;        for(k=fail[i-1];k>=0&&pat[i]!=pat[k+1];k=fail[k]);        if(pat[k+1]==pat[i]) fail[i]=k+1;    }  //  for(int i=0;pat[i];i++) cout<<fail[i]<<endl;}int main(){    int n,cas=0;    while(~scanf("%d",&n)&&n){        scanf("%s",str);        get_next(str);        int len=strlen(str);        printf("Test case #%d\n",++cas);        for(int i=0;i<len;i++){            int plen=i-fail[i];            if(i+1!=plen&&(i+1)%plen==0){                printf("%d %d\n",i+1,(i/plen)+1);            }        }        printf("\n");    }    return 0;}