POJ 1961-Period （KMP）

Period

Time Limit: 1000ms   Memory limit: 65536K  有疑问？点这里^_^

题目描述

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 AK ,

that is A concatenated K times, for some string A. Of course, we also want to

know the period K.

 

输入

 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.

 

输出

 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.

 

示例输入

3aaa12aabaabaabaab0

示例输出

Test case #12 23 3Test case #22 26 29 312 4

提示

 

来源

 Southeastern European Regional Programming Contest

 

示例程序

 

KMP的简单运用。求字符串的最大子串。

#include <stdio.h>char s[1000010];int next[1000010],n;int getnext( ){    int i=0,j=-1;    next[0]=-1;    while(i<n)    {        if(j==-1||s[i]==s[j])        {            i++;            j++;            next[i]=j;        }        else            j=next[j];    }}int main( ){    int i;    int k=0;    while(scanf("%d",&n)!=EOF)    {        if(n==0)            break;        k++;        scanf("%s",s);        getnext( );        printf("Test case #%d\n",k);        for( i=2; i<=n; i++)        {            if(i%(i-next[i])==0)            {                if(i/(i-next[i])>1)                    printf("%d %d\n",i,i/(i-next[i]));            }        }        printf("\n");    }}