HDU 1358 Period 求前缀长度和出现次数（KMP的next数组的使用）

Period

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12428    Accepted Submission(s): 5825

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

 

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题目大意：
给一个字符串，从第二个字符开始，让你判断前面字符串是否具有周期性，然后输出此位置和最大周期数。（周期要大于一）

 

思路：

先构造出 next[] 数组，下标为 i，定义一个变量 j = i - next[i] 就是next数组下标和下标对应值的差，如果这个差能整除下标 i，即 i%j==0 ,则说明下标i之前的字符串（周期性字符串长度为 i）一定可以由一个前缀周期性的表示出来，这个前缀的长度为刚才求得的那个差，即 j，则这个前缀出现的次数为 i/j 。所以最后输出i和i/j即可。

 

code：

 

#include<cstdio>
#include<iostream>
#include<cstring>
#include<memory>
using namespace std;
char wenben[1000005];
int next1[1000005];
void getnext1(char* s,int* next1,int m)
{
    next1[0]=0;
    next1[1]=0;
    for(int i=1;i<m;i++)
    {
        int j=next1[i];
        while(j&&s[i]!=s[j])
            j=next1[j];
        if(s[i]==s[j])
            next1[i+1]=j+1;
        else
            next1[i+1]=0;
    }
}
int main()
{
    int L;
    int t=1;
    while(~scanf("%d",&L))
    {
        if(L==0)
            break;
        scanf("%s",wenben);
        getnext1(wenben,next1,L);
        for(int i=0;i<L;i++)
            printf("i=%d->%d ",i,next1[i]);
        printf("
");
        printf("Test case #%d
",t++);
        for(int i=2;i<=L;i++)
        {
            int k=i-(next1[i]);
            if(k!=i&&(i)%k==0)
                printf("%d %d
",i,i/k);
        }
        printf("
");
    }
    return 0;
}