HDU 1358 Period

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Period

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5098 Accepted Submission(s): 2467

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

3aaa12aabaabaabaab0

Sample Output

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

这个因为不是匹配字符串，所以不需要当x[i] == x[i + 1]时令两者next的值相等。而这个题又是求重复的循环子串的个数的问题，根据之前做的求循环子串补齐大小的题目类似。我们只需要确认只要当前位置与当前位置的next数组的值的差是当前位置的一个因子即可，且必须要求这个因子不能是当前位置本身。这个很好证明，当前位置与当前位置的next数组的值的差是循环子串的长度，只要这个长度是一个因子，当然是满足是循环的。

代码如下：

/*************************************************************************> File Name: Period.cpp> Author: Zhanghaoran> Mail: chilumanxi@xiyoulinux.org> Created Time: Tue 24 Nov 2015 09:01:57 PM CST ************************************************************************/#include <iostream>#include <algorithm>#include <cstring>#include <cstdio>#include <cstdlib>using namespace std;void preKMP(char x[], int m, int nextKMP[]){    int i , j;    j = nextKMP[0] = -1;    i = 0;    while(i <= m){        if(j == -1 || x[i] == x[j]){            nextKMP[++ i] = ++ j;            if(i % (i - j) == 0 && i / (i - j) > 1){                cout << i << " " << i / (i - j) << endl;            }        }        else             j = nextKMP[j];    }    cout << endl;}int nexti[1000010];void KMP_Count(char x[], int m){    int i, j;    int ans = 0;    preKMP(x, m, nexti);}int N;char a[1000010];int main(void){    int cas = 1;    while(1){        scanf("%d", &N);        if(N == 0)            break;        printf("Test case #%d\n",cas ++);        scanf("%s", a);           KMP_Count(a, N);    }}

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