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Period
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
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 AK , 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
题目大意：
对于输入的字符串，找出串中所有的子串（保证该字符串由这个子串组成，并且长度大于1）
输出子串的长度和组成这个字符串所需要的这个子串的个数
解题思路：
KMP算法 next数组的应用
先找出 循环节（长度大于1），然后判断输入的字符串能否由该循环节组成。
对KMP不太熟悉的可以看这里:从头到尾理解KMP算法
1 #include <stdio.h> 2 #include <string.h> 3 4 char s[1000010]; 5 int next_[1000010]; 6 void getnext(){ 7 int i = 0, j = -1; 8 int len = strlen(s); 9 next_[0] = -1; 10 while (i < len){ 11 if (j == -1 || s[i] == s[j]) 12 next_[++ i] = ++ j; 13 else 14 j = next_[j]; 15 } 16 } 17 int main () 18 { 19 int n,i,j,f = 1; 20 while (~scanf("%d",&n),n) 21 { 22 scanf("%s",s); 23 getnext(); 24 printf("Test case #%d\n",f ++); 25 for (i = 1; i <= n; i ++) 26 if (next_[i]) // i-next_[i] 为循环节的长度 27 if (i%(i-next_[i])==0 && i/(i-next_[i])!=1) //判断输入的字符串能否由该循环节组成 28 printf("%d %d\n",i,i/(i-next_[i])); 29 printf("\n"); 30 } 31 return 0; 32 }
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