Period||POJ1961

最新推荐文章于 2021-07-19 17:07:24 发布

原创最新推荐文章于 2021-07-19 17:07:24 发布 · 218 阅读

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本文介绍了一种使用KMP算法解决字符串问题的方法，具体为判断一个字符串的每个前缀是否具有周期性，并找到该周期的长度。通过构建next数组进行匹配，实现高效求解。

link:http://poj.org/problem?id=1961
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 AK ,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

3
aaa
12
aabaabaabaab
0

Sample Output

Test case #1
2 2
3 3

Test case #2
2 2
6 2
9 3
12 4

题解：这个和poj2406基本相似
可以看看这个http://blog.youkuaiyun.com/csdn_muxin/article/details/77231807
kmp简单运用

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<iostream>
using namespace std;
char s[1000100];
int next[1000100];
void kmp()
{
    int n=strlen(s);
    int i,j;
    i=0,j=-1;
    next[0]=-1;
    while(s[i])
    {
        if(j==-1||s[i]==s[j])
        {
            i++,j++;
            next[i]=j;
        }
        else
            j=next[j];
    }
}
int main()
{
    int n,i,j,c=1;
    while(scanf("%d",&n),n!=0)
    {
        scanf("%s",s);
        kmp();
        printf("Test case #%d\n",c++);
        for(i=1;i<=n;i++)
        {
            if(i%(i-next[i])==0&&i/(i-next[i])>1)
            {
                printf("%d %d\n",i,i/(i-next[i]));
            }
        }
        printf("\n");
    }
return 0;
}