Period
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 8107 Accepted Submission(s): 3903
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
分析：1 题目要求的是给定一个长度为n的字符串，求出字符串的所有前缀字符串中该字符串刚好由k个最小循环节够成，由于题目要求k最大那么就是这个循环节最小2 只要先求出next数组，然后去枚举所有的前缀找到满足的输出长度和k
#include<algorithm>
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define MAXN 1000010
int n;
int Next[MAXN];
char pattern[MAXN];
void getNext(char *str){
int m = strlen(str);
Next[0] = Next[1] = 0;
for(int i = 1 ; i < m ; i++){
int j = Next[i];
while(j && str[i] != str[j])
j = Next[j];
Next[i+1] = str[i] == str[j] ? j+1 : 0;
}
}
int main(){
int Case = 1;
while(scanf("%d%*c" , &n) && n){
scanf("%s" , pattern);
getNext(pattern);
printf("Test case #%d\n" , Case++);
for(int i = 2 ; i <= n ; i++){
int len = i;
int cir = len-Next[len];
if(cir == 1)/*如果最小循环节为1那么肯定是满足的*/
printf("%d %d\n" , len , len);
else if(cir != len && len%cir == 0)/*这里注意一下是cir != len*/
printf("%d %d\n" , len , len/cir);
else
continue;
}
printf("\n");
}
return 0;
}
