hdu 1358(KMP--next[]数组应用
标签：div string \n 技术分享 ber hid -- targe for
Period
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13060 Accepted Submission(s): 6102
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 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
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
Sample Output
Test case #1
2 2
3 3
Test case #2
2 2
6 2
9 3
12 4
1 /**
2 Finished time:9.15.2018
3 Author : hyacinth
4 Algorithm : KMP--next[]
5 */
6 #include 7 #include 8 #include 9 using namespace std;
10 const int maxn=1000010;
11
12 int nex[maxn]={-1,0};
13 char buf[maxn];
14
15 void Get_Next(const char s[])
16 {
17 int len=strlen(s);
18 int index=2;
19 int cn=0;
20 while(index len)
21 {
22 if(s[index-1] == s[cn])
23 nex[index++]=++cn;
24 else if(nex[cn] != -1)
25 cn=nex[cn];
26 else
27 nex[index++]=0;
28 }
29 }
30 int main()
31 {
32 int N;
33 int kase=1;
34 while(~scanf("%d",&N) && N)
35 {
36 scanf("%s",buf);
37 Get_Next(buf);
38 int len=strlen(buf);
39 printf("Test case #%d\n",kase++);
40 for(int i=2;i i)
41 {
42 int ans=i-nex[i];
43 int k=i/ans;
44 if(i%ans == 0 && k > 1)
45 printf("%d %d\n",i,k);
46 }
47 printf("\n");
48 }
49 }
View Code
hdu 1358(KMP--next[]数组应用
标签：div string \n 技术分享 ber hid -- targe for
原文地址：https://www.cnblogs.com/violet-acmer/p/9650785.html
评论