HDU1358 Period

最新推荐文章于 2020-02-07 11:05:28 发布

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Period

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2766 Accepted Submission(s): 1368

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

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

题意：输入一串长度为n的字符串，要求求出每个前缀的最短循环节。。
分析：通过这个题目对KMP的next更加熟悉了，KMP的next数组就是i位置的后缀字符串和他的前缀字符串相等的最长长度。比如aaba，对应位置是1,2,3,4,若果当前位置是3，第一个位置的a不等于第三个位置的c，所以next[3]=0;位置是4的话，第四个位置的a与第一个位置的a相等，所以next[4]=1;注意next存放的是最大的匹配长度，如果长度不为1，那么匹配就是t1,t2......t(k)与t(i-k+1),t(i-k+2).......t(i)。相等，k就是next里的值。。求出next之后i-next表示i和next[i]之间有多少个字母，这就是循环节

#include<cstdio>
#include<cstring>
using namespace std;
const int MAXN=1000010;
char s[MAXN];
int n;
int next[MAXN];
int main()
{
    int i,flag=1;
    while(scanf("%d",&n)==1&&n)
    {
        scanf("%s",s+1);
        next[1]=0;
        int k=0;
        for(i=2;i<=n;i++)
        {
            while(k&&s[i]!=s[k+1])
                k=next[k];
            if(s[k+1]==s[i])
                k++;
            next[i]=k;
        }
        printf("Test case #%d\n",flag++);
        for(i=2;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;
}