Power Strings
| Time Limit: 3000MS | Memory Limit: 65536K | |
| Total Submissions: 30069 | Accepted: 12553 |
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
KMP算法,next[]数组的理解。
这道题考察的是对next[]数组的理解。如果 i 能整除 (i-next[i]) 的话，证明在这个字符之前，是一个由同一前缀重复连接构成的字符串。例如：aabaabaabc，c位置的下标 i 就可以整除他的下标和next数组值的差(i-next[i])，则它之前的字符串aabaabaab就是一个重复字符串，构成它的前缀的长度就是 (i - next[i])。
注意：输入以一个'.'结束；注意这一组测试数据“aabaabaa”，WA的可以测试一下，正确结果应该为1。
如果觉得文字太枯燥可以猛戳这里：hdu 1358:Period（KMP算法，next[]数组的使用）
代码：
#include <stdio.h> char s[];
int next[];
void GetNext(char a[],int next[],int n) //获得a数列的next数组
{
int i=,k=-;
next[] = -;
while(i<n){
if(k==-){
next[i+] = ;
i++;k++;
}
else if(a[i]==a[k]){
next[i+] = k+;
i++;k++;
}
else
k = next[k];
}
} int main()
{
while(scanf("%s",s)!=EOF){
if(s[]=='.') break;
int i;
for(i=;s[i];i++);
int len = i;
GetNext(s,next,len);
if(len%(len-next[len])==)
printf("%d\n",len/(len-next[len]));
else
printf("1\n");
}
return ;
}
Freecode : www.cnblogs.com/yym2013