poj--1961--Period
来源:互联网 发布:网络综艺为什么这么火 编辑:程序博客网 时间:2024/06/08 17:38
Period
Time Limit: 3000MS Memory Limit: 30000KTotal Submissions: 18603 Accepted: 9044
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.
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
3aaa12aabaabaabaab0
Sample Output
Test case #12 23 3Test case #22 26 29 312 4
题意:
求字符串的每个前缀的大于等于2的循环节长度
思路:
考察KMP算法中对next数组的理解,如果对KMP算法比较了解的话,很容易就能写出来
代码:
C++ Code
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
#include<cstdio>
using namespace std;
const int MAXN = 1000010;
char str[MAXN];
int next[MAXN];
int n;
void getNext()
{
int i, j;
i = 0;
j = -1;
next[0] = -1;
while(str[i] != '\0')
{
if(j == -1 || str[i] == str[j])
{
i++;
j++;
if(i % (i - j) == 0 && i / (i - j) > 1)
printf("%d %d\n", i, i / (i - j));
next[i] = j;
}
else j = next[j];
}
}
int main()
{
int iCase = 0;
while(scanf("%d", &n), n)
{
iCase++;
scanf("%s", &str);
printf("Test case #%d\n", iCase);
getNext();
printf("\n");
}
return 0;
}
using namespace std;
const int MAXN = 1000010;
char str[MAXN];
int next[MAXN];
int n;
void getNext()
{
int i, j;
i = 0;
j = -1;
next[0] = -1;
while(str[i] != '\0')
{
if(j == -1 || str[i] == str[j])
{
i++;
j++;
if(i % (i - j) == 0 && i / (i - j) > 1)
printf("%d %d\n", i, i / (i - j));
next[i] = j;
}
else j = next[j];
}
}
int main()
{
int iCase = 0;
while(scanf("%d", &n), n)
{
iCase++;
scanf("%s", &str);
printf("Test case #%d\n", iCase);
getNext();
printf("\n");
}
return 0;
}
阅读全文
0 0
- POJ 1961 Period
- poj 1961 Period
- POJ 1961 Period
- poj 1961 Period
- poj 1961:Period (KMP)
- POJ-1961 Period
- poj 1961 Period
- POJ 1961 - Period
- poj 1961 Period
- poj 1961Period
- poj 1961 Period (KMP)
- POJ-1961-Period
- POJ 1961 Period
- poj 1961 Period
- poj 1961 Period
- poj 1961 Period---kmp
- POJ 1961 Period
- poj-Period-1961
- linux下使用润乾V5设计器乱码问题
- eval解析JSON中的注意点 在JS中将JSON的字符串解析成JSON数据格式,一般有两种方式: 1.一种为使用eval()函数。 2. 使用Function对象来进行返回解析。 使用eval
- 计算机修炼之路--------JavaScript法术的学习笔记(三)之JavaScript语法(二)
- hdu6138 多校2017 ac自动机or后缀数组
- Debug Knowledge Base
- poj--1961--Period
- C/C++的传指针和引用
- HDU-4847:Wow! Such Doge!(震惊!!!kmp模板题。。。kmp:我不要面子的啊)
- Jmeter中Websocket协议支持包的使用
- codeforces734D
- tensorflow基础
- 只用路径上传文件,不用手动选择文件上传
- Dipping into Shared Memory
- Super Jumping! Jumping! Jumping! hdu1087