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NC50990. Period

描述

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 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.

输入描述

The input consists of several test cases. Each test case consists of two lines. The first one contains N – 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.

输出描述

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.

示例1

输入:

3
aaa
12
aabaabaabaab
0

输出:

Test case #1
2 2
3 3

Test case #2
2 2
6 2
9 3
12 4

原站题解

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C++14(g++5.4) 解法, 执行用时: 51ms, 内存消耗: 8276K, 提交时间: 2020-06-06 10:19:36

#include<bits/stdc++.h>
using namespace std;
const int N=1e6+10;
char str[N];
int net[N],n;
void get_next()
{
	
	for(int i=2,j=0;i<=n;i++)
	{
	   while(j>0&&str[i]!=str[j+1])j=net[j];
	   if(str[i]==str[j+1])j++;
	   net[i]=j;
	}
}
int main()
{
	int T=0;
	while(scanf("%d",&n),n)
	{
		scanf("%s",str+1);
		get_next();
		printf("Test case #%d\n",++T);
		for(int i=2;i<=n;i++)
		{
			int t=i-net[i];
			if(i!=t&&i%t==0)printf("%d %d\n",i,i/t);
		}
		puts("");
	}
 }

C++(g++ 7.5.0) 解法, 执行用时: 43ms, 内存消耗: 10272K, 提交时间: 2023-07-27 10:02:36

#include<bits/stdc++.h>
using namespace std;
char a[1000001];
long long nex[1000001],n,t;
inline void calc_next(){
	nex[1]=0;
	for(register int i=2,j=0;i<=n;++i){
		while(j&&a[i]!=a[j+1])j=nex[j];
		if(a[i]==a[j+1])++j;
		nex[i]=j;
	}
}
int main(){
	while(scanf("%lld",&n)&&n){
		scanf("%s",a+1);calc_next();printf("Test case #%d\n",++t);
		for(register int i=2;i<=n;++i)if(i%(i-nex[i])==0&&i/(i-nex[i])>1)printf("%lld %lld\n",i,i/(i-nex[i]));
		puts("");
	}
}

C++(clang++ 11.0.1) 解法, 执行用时: 39ms, 内存消耗: 6356K, 提交时间: 2023-03-19 08:28:13

#include<iostream>
using namespace std;
char s[10000000];
int n,ne[10000000],T=1;
int main()
{
    while(cin>>n,n)
    {
		scanf("%s",s+1);
		for(int i=2,j=0;i<=n;i++)
		{
			while(j>0&&s[i]!=s[j+1])j=ne[j];
			if(s[i]==s[j+1])j++;
			ne[i]=j;
		}
		printf("Test case #%d\n",T++);
		for(int i=1;i<=n;i++)
		{
			int t=i-ne[i];
			if(i%t==0&&i/t>1)
				printf("%d %d\n",i,i/t);
		}
		puts("");
	}
}

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