POJ Power Strings 2406【KMP】
2015-08-07 12:06
369 查看
Power Strings
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
Sample Output
Hint
This problem has huge input, use scanf instead of cin to avoid time limit exceed.
Source
Waterloo local 2002.07.01
Time Limit: 3000MS | Memory Limit: 65536K | |
Total Submissions: 37379 | Accepted: 15443 |
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
#include <stdio.h> #include <string.h> #include <algorithm> #define maxn 1000000+10 using namespace std; char P[maxn]; int pre[maxn]; void getnext(int plen) { pre[0]=pre[1]=0; for(int i=1;i<plen;i++){ int k=pre[i]; while(k&&P[i]!=P[k])k=pre[k]; pre[i+1]= P[k]==P[i]?k+1:0; } } int main() { while(scanf("%s",P)!=EOF){ if(P[0]=='.')break; int plen=strlen(P); getnext(plen); if(plen%(plen-pre[plen])==0) printf("%d\n",plen/(plen-pre[plen])); else printf("1\n"); } return 0; }
相关文章推荐
- mysql rand()产生随机整数范围及方法
- 有道词典在谷歌Chrome浏览器中无法取词的解决方法 分类: 开发工具 2015-08-07 12:06 7人阅读 评论(0) 收藏
- Nodejs Http发送post请求
- C++ list
- leetcode3:不重复的最长子串长度
- 【ASP.NET】视频总结
- e2fsck
- 测绘&遥感&地信 国内外期刊大全
- java线程研究---(2)启动Thread
- 2015-08-07 OJ初级练习
- leetcode2:Add Two Numbers
- 欢迎使用CSDN-markdown编辑器
- C++ 记录程序运行时间
- Mysql注入总结(一)
- 修改mysql密码
- Unity随机Prefab,自动前往某点处理
- 暑期学校ACM之旅
- Unity随机Prefab,自动前往某点处理
- 搜索引擎与SEO的纽带——你应该知道的事:网络爬虫
- jQuery手机端上拉刷新下拉加载更多页面