poj2406 Power Strings(kmp失配函数)
Power Strings
Time Limit: 3000MS |
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Memory Limit: 65536K |
Total Submissions: 39291 |
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Accepted: 16315 |
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
【思路】
KMP。
应用KMP算法中的失配函数,如果是一个周期串那么错位部分(n-f[n])一定是一个最小循环节(仔细想想f函数的意义),则答案为i/(n-f[n])。否则ans=1。
时间复杂度为O(n)。
【代码】
1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #define FOR(a,b,c) for(int a=(b);a<=(c);a++) 5 using namespace std; 6 7 const int maxn = 1000000+10; 8 9 void getFail(char* P,int* f) { 10 int m=strlen(P); 11 f[0]=f[1]=0; 12 for(int i=1;i<m;i++) { 13 int j=f[i]; 14 while(j && P[i]!=P[j]) j=f[j]; 15 f[i+1]=P[i]==P[j]?j+1:0; 16 } 17 } 18 19 char s[maxn]; 20 int f[maxn]; 21 22 int main() { 23 while(scanf("%s",s)==1 && s[0]!='.') { 24 getFail(s,f); 25 int n=strlen(s); 26 if(f[n]>0 && n%(n-f[n])==0) printf("%d\n",n/(n-f[n])); 27 else printf("%d\n",1); 28 } 29 return 0; 30 }
posted on 2015-11-28 17:44 hahalidaxin 阅读(376) 评论(0) 编辑 收藏 举报