poj 1961 KMP算法的应用

最新推荐文章于 2021-09-01 21:48:51 发布

最新推荐文章于 2021-09-01 21:48:51 发布 · 68 阅读

博主分享了解决杭电在线评测系统(OJ)中一道关于字符串周期性问题的心得。该题要求对于给定的字符串，找出所有前缀是否为周期性字符串，并给出周期长度。博主使用KMP算法中的next数组来解决此问题。

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多久了，，，多久了，，，，两天了，，，，终于ac了一道题，，还是一道水题。。。。。。悲剧，，，，杭电上的一道题，，，一直看不懂题意，看不懂，，纠结，，，，就这样一直看了两天，，还是不懂，，，受打击了，，浪费了两天时间，，杭电3336，，，求高手指导。。。题目：

Period

Time Limit:3000MS		Memory Limit:30000K
Total Submissions:8387		Accepted:3734

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 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

ac代码：

#include <iostream> #include <cstdio> #include <string.h> using namespace std; const int N=1000010; int nextt[N]; char str[N]; int len; void get_nextt(){ nextt[0]=0; int i,temp; for(i=1;i<len;++i){ temp=nextt[i-1]; while(temp&&str[temp]!=str[i]) temp=nextt[temp-1]; if(str[temp]==str[i]) nextt[i]=temp+1; else nextt[i]=0; } } int main(){ int num=0; while(scanf("%d",&len)&&len){ scanf("%s",str); get_nextt(); printf("Test case #%d\n",++num); for(int i=1;i<len;i++){ if(nextt[i]==0) continue; int x=(i+1)-nextt[i]; if(((i+1)%x==0)) printf("%d %d\n",i+1,(i+1)/x); } printf("\n"); } return 0; }