POJ 1961:Period
2015-06-27 13:46
351 查看
Period
Time Limit: 3000MS Memory Limit: 30000K
Total Submissions: 14280 Accepted: 6773
Description
For each prefix of a given string S with N characters (each character has an ASCII code between 97 and 126, inclusive), we want to know whether the prefix is a periodic string. That is, for each i (2 <= i <= N) we want to know the largest K > 1 (if there is one) such that the prefix of S with length i can be written as AK ,that is A concatenated K times, for some string A. Of course, we also want to know the period K.
Input
The input consists of several test cases. Each test case consists of two lines. The first one contains N (2 <= N <= 1 000 000) – the size of the string S.The second line contains the string S. The input file ends with a line, having the
number zero on it.
Output
For each test case, output “Test case #” and the consecutive test case number on a single line; then, for each prefix with length i that has a period K > 1, output the prefix size i and the period K separated by a single space; the prefix sizes must be in increasing order. Print a blank line after each test case.
Sample Input
3
aaa
12
aabaabaabaab
0
Sample Output
Test case #1
2 2
3 3
Test case #2
2 2
6 2
9 3
12 4
POJ2406与这道题一个意思,就是这道题细化了一点。
代码:
Time Limit: 3000MS Memory Limit: 30000K
Total Submissions: 14280 Accepted: 6773
Description
For each prefix of a given string S with N characters (each character has an ASCII code between 97 and 126, inclusive), we want to know whether the prefix is a periodic string. That is, for each i (2 <= i <= N) we want to know the largest K > 1 (if there is one) such that the prefix of S with length i can be written as AK ,that is A concatenated K times, for some string A. Of course, we also want to know the period K.
Input
The input consists of several test cases. Each test case consists of two lines. The first one contains N (2 <= N <= 1 000 000) – the size of the string S.The second line contains the string S. The input file ends with a line, having the
number zero on it.
Output
For each test case, output “Test case #” and the consecutive test case number on a single line; then, for each prefix with length i that has a period K > 1, output the prefix size i and the period K separated by a single space; the prefix sizes must be in increasing order. Print a blank line after each test case.
Sample Input
3
aaa
12
aabaabaabaab
0
Sample Output
Test case #1
2 2
3 3
Test case #2
2 2
6 2
9 3
12 4
POJ2406与这道题一个意思,就是这道题细化了一点。
代码:
#include <iostream> #include <vector> #include <string> #include <cstring> #include <algorithm> using namespace std; char a[1000005]; int next1[1000005]; void cal() { int len = strlen(a); int i,j=-1; next1[0]=-1; for(i=0;i<len;) { if(j==-1||a[i]==a[j]) { i++; j++; next1[i]=j; } else { j=next1[j]; } } } int main() { int len,count=1; while(cin>>len) { if(len==0) break; cin>>a; cal(); int i; cout<<"Test case #"<<count++<<endl; for(i=2;i<=len;i++) { if(i%(i-next1[i])==0 && i/(i-next1[i])>=2) cout<<i<<" "<<i/(i-next1[i])<<endl; } cout<<endl; } return 0; }
相关文章推荐
- Android自定义键盘
- 指针专题
- 三种快速排序算法以及快速排序的优化
- DMALL刘江峰:生鲜市场具有巨大O2O改造空间
- Linux下的一个图形管理工具webmin
- 移动前端—图片压缩上传实践
- samba
- grid++report中篇
- SDK Manager打不开
- js——DOM操作(二)
- 面向对象基础
- django admin 根据用户显示不同的列表以及编辑界面等
- Redis 笔记与总结5 Redis 常用命令之 键值命令 和 服务器命令 && 高级应用之 安全性 和 主从复制
- centos device eth0 does not seem to be present解决方法
- $GLOBALS['HTTP_RAW_POST_DATA'] 和$_POST的区别
- 设计模式——简单工厂模型
- sgu 246 分类: sgu 2015-06...
- sgu 246
- 【转】三种快速排序算法以及快速排序的优化
- sgu 246 分类: sgu 2015-06-27 13:40 20人阅读 评论(0) 收藏