HDU - 1358 Period_hdu period-优快云博客

本文深入探讨了代码逻辑的解析与优化技术，通过实际案例分析，详细讲解了如何提高代码效率和性能，包括使用现代编程语言特性、优化算法、资源管理策略等关键点。此外，还分享了常见代码问题的解决思路与最佳实践，旨在帮助开发者提升编程技能，实现高效、稳定的软件开发。

Period

Time Limit: 1000MS

Memory Limit: 32768KB

64bit IO Format: %I64d & %I64u

[Submit] [Go Back] [Status]

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 A ^K , that is A concatenated K times, for some string A. Of course, we also want to know the period K.

Input

The input file 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

Source

kmp求周期，简单题：

//本题的输入排除了周期i/k=1的情况
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn=1000000+10;
int next[maxn];
char s[maxn];
int n;
void getnext(char *s)//kmp不仅出现在kmp函数，在其他函数也都用到了kmp
{
int j=-1,i=0;
next[0]=-1;
while(i<n)
{
if(j==-1||s[i]==s[j])
{
i++,j++;
next[i]=j;//next[i]表示i位置的上一个位置（即i-1位置）的部分匹配值
}
//else j=-1;//还是这句简洁高效，错！
else j=next[j];//为了防止出错，用这句更保险（需要字符串自身匹配，只能用这句，不能用楼上）
}
}

int main()
{
int i,t=0;
while(scanf("%d",&n)&&n!=0)
{
t++;
scanf("%s",s);
getnext(s);
printf("Test case #%d\n",t);
if(n==2) printf("2 2\n\n");//严格按照题意，这里要加上这句话，否则当n=2，只输出Test case #t。而题意是允许n=2的。但本题不写这句也可以AC
else
{
for(i=2;i<=n;i++)
{
int k=i-next[i];//题目要求周期i/k>1
if(i/k>1&&i%k==0) printf("%d %d\n",i,i/k);
}
printf("\n");
}
}
return 0;
}