題干：

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

?

題目大意：

? ? 給你一個字符串，求這個字符串到第i個字符為止的子字符串的循環節的次數，輸出i和循環次數。

解題報告：

? ?運用next數組可以直接出結果、、

AC代碼：

#include<cstdio> #include<iostream> #include<algorithm> #include<queue> #include<map> #include<vector> #include<set> #include<string> #include<cmath> #include<cstring> #define ll long long #define pb push_back #define pm make_pair #define fi first #define se second using namespace std; const int MAX = 2e6 + 5; char s[MAX]; int Next[MAX],len; void getnext() {Next[0] = -1;int k = -1,j = 0;int len = strlen(s);while(j < len) {if(k == -1 || s[k] == s[j]) {k++,j++;Next[j] = k;} else k = Next[k];} } int main() {int n,iCase = 0;while(~scanf("%d",&n)) {if(n == 0) break;scanf("%s",s);getnext();printf("Test case #%d\n",++iCase);for(int i=2; i<=n; i++) {int tmplen=i-Next[i]; //循環節長度if(Next[i]>0 && i % tmplen == 0) printf("%d %d\n",i,i/tmplen);}printf("\n");}return 0; }

?

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