记录编号 |
364633 |
评测结果 |
AAAAAAAAAA |
题目名称 |
[POJ2406]字符串的幂 |
最终得分 |
100 |
用户昵称 |
ONCE AGAIN |
是否通过 |
通过 |
代码语言 |
C++ |
运行时间 |
0.192 s |
提交时间 |
2017-01-17 15:16:57 |
内存使用 |
5.06 MiB |
显示代码纯文本
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <ctime>
using namespace std;
const int maxn = 1000010;
char ch[maxn] = "\0";
int fail[maxn] = {0};
void GetFail(int n){
int j = 0;fail[1] = 0;
for(int i = 2;i <= n;i++){
while(j>0&&ch[j+1]!=ch[i])j=fail[j];
if(ch[i]==ch[j+1])j++;
fail[i] = j;
}
}
int main(){
freopen("powerstrings.in","r",stdin);freopen("powerstrings.out","w",stdout);
while(scanf("%s",ch+1)!=EOF&&ch[1]!='.'){
int len = strlen(ch+1);
GetFail(len);
if(len - fail[len]!=0&&len % (len - fail[len]) == 0)printf("%d\n",len/(len - fail[len]));
else printf("1\n");
}
return 0;
}