【KMPnxt数组应用】POJ2406Power Strings

Power Strings
Time Limit: 3000MS        Memory Limit: 65536K
Total Submissions: 60872        Accepted: 25166
Description

Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b = "abcdef". If we think of concatenation as multiplication, exponentiation by a non-negative integer is defined in the normal way: a^0 = "" (the empty string) and a^(n+1) = a*(a^n).
Input

Each test case is a line of input representing s, a string of printable characters. The length of s will be at least 1 and will not exceed 1 million characters. A line containing a period follows the last test case.
Output

For each s you should print the largest n such that s = a^n for some string a.
Sample Input

abcd
aaaa
ababab
.
Sample Output

1
4
3
Hint

This problem has huge input, use scanf instead of cin to avoid time limit exceed.
Source

Waterloo local 2002.07.01

T

这道题就是求循环节的长度

也就是个nxt数组应用

答案就是n/(n-nxt[n])，注意n%(n-nxt[n])为真的时候要输出1

这个是为什么呢

动手画一画，nxt数组的特性就是这样，n-nxt[n]恰好就是一个循环节的长度

因为我是习惯全部+1来写，所以就很玄学

有一个最好要思考的细节

为什么是nxt[n+1]，因为在n+1的时候才会失配（正常的KMP应该是0到len-1，然后nxt[len]是答案）

 1 #include<cstdio>
 2 #include<algorithm>
 3 #include<cstring>
 4 using namespace std;
 5 char a[1000000+100];
 6 int nxt[1000000+100],alen;
 7 void getnxt()
 8 {
 9     int j=1,k=0;
10     nxt[1]=0;
11     while(j<=alen)
12     {
13         if(k==0||a[j]==a[k])
14             j++,k++,nxt[j]=k;
15         else
16             k=nxt[k];
17     }
18 }
19 int main()
20 {
21     while(1)
22     {
23         memset(nxt,0,sizeof(nxt));
24         scanf("%s",a+1);
25         if(a[1]=='.')return 0;
26         alen=strlen(a+1);
27         getnxt();
28         if(alen%(alen+1-nxt[alen+1]))puts("1"); 
29         else printf("%d\n",(alen)/((alen+1)-nxt[alen+1]));
30     }
31     return 0;
32 }

 

转载于:https://www.cnblogs.com/Qin-Wei-Kai/p/10205698.html