[POJ2406] Power Strings

目录

• Description
• Input
• Output
• Sample Input
• Sample Output
• Hint
• 题解
  • 暴力
  • $KMP$
  • 超时的$Lcp$
  • 字符串$Hash$

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

abcd
aaaa
ababab
.

Sample Output

1
4
3

Hint

This problem has huge input, use scanf instead of cin to avoid time limit exceed.

题解

(PS:)这题的正解是(KMP)算法，这里还给出超时的(Lcp)与侥幸过了的暴力

---

暴力

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;

const int L=1001001;
int l,id; char s[L];

int Check()
{
	for(int i=id;i<l;++i)
		if(s[i]!=s[i-id]) return 0;
	return 1;
}

int main()
{
	int Ans;
	for(scanf("%s",s);s[0]!='.';scanf("%s",s))
	{
		l=strlen(s);
		Ans=1;
		for(int i=l;i>=2;--i)
		{
			if(l%i) continue;
			id=l/i;
			if(Check()) {Ans=i;break;}
		}
		printf("%d
",Ans);
	}
	return 0;
}

---

(KMP)

(KMP)的巧妙之处在于它的(next)数组，该数组能反映字符串的对称性

这题还有一个坑点，要加上特判

if(l%(l-nt[l])==0) Ans=l/(l-nt[l]);

即可以分解为整数个子串

(My~Code)

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;

const int L=1001001;
int l,nt[L];
char s[L];

void Make_nt()
{
	int i=0,j=-1;
	nt[i]=-1;
	while(i<l)
	{
		if(j==-1||s[i]==s[j])
			++i,++j,nt[i]=j;
		else j=nt[j];
	}
}

int main()
{
	int len,Ans;
	for(scanf("%s",s);s[0]!='.';scanf("%s",s))
	{
		l=strlen(s);
		Make_nt();
		Ans=1;
		if(l%(l-nt[l])==0) Ans=l/(l-nt[l]);
		printf("%d
",Ans);
	}
	return 0;
}

---

超时的(Lcp)

(My~Code)

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;

const int N=1001001;
int n,sa[N],rk[N],tmp[N],c[N],height[N],Qz[N];
char s[N];

void Make_Sa()
{
	int i,j,len,na;
	na=256;
	memset(c,0,na*sizeof(int));
	for(i=0;i<n;++i)
	{
		rk[i]=s[i]&0xff,
		++c[rk[i]];
	}
	for(i=1;i<na;++i) c[i]+=c[i-1];
	for(i=0;i<n;++i)
	{
		--c[rk[i]],
		sa[c[rk[i]]]=i;
	}
	for(len=1;len<n;len<<=1)
	{
		for(i=0;i<n;++i)
		{
			j=sa[i]-len;
			if(j<0) j+=n;
			tmp[c[rk[j]]++]=j;
		}
		c[0]=0,j=0,
		sa[tmp[0]]=0;
		for(i=1;i<n;++i)
		{
			if(rk[tmp[i]]!=rk[tmp[i-1]]
			||rk[tmp[i]+len]!=rk[tmp[i-1]+len])
				c[++j]=i;
			sa[tmp[i]]=j;
		}
		memcpy(rk,sa,n*sizeof(int)),
		memcpy(sa,tmp,n*sizeof(int));
		if(j>=n-1) break;
	}
}

void Lcp()
{
	int i=0,j,k=0;
	height[0]=0;
	for(j=rk[0];i<n-1;++i,++k)
		for(;k>=0&&s[i]!=s[sa[j-1]+k];)
		{
			height[j]=k, --k;
			j=rk[sa[j]+1];
		}
}

void Solve()
{
	Qz[rk[0]]=n-1;
	for(int i=rk[0]-1;i;--i) Qz[i]=min(Qz[i+1],height[i+1]);
	for(int i=rk[0]+1;i<n;++i) Qz[i]=min(Qz[i-1],height[i-1]);
	int Ans=-1;
	--n;
	for(int i=1;i<=n;++i)
	{
		if(n%i) continue;
		if(Qz[rk[i]]==n-i)
		{
			Ans=n/i;
			break;
		}
	}
	printf("%d
",Ans);
}

int main()
{
	for(;;)
	{
		scanf("%s",s);
		n=strlen(s);
		if(n==1&&s[0]=='.') break;
		s[n++]=1;
		Make_Sa(),Lcp(),Solve();
	}
	return 0;
}

---

字符串(Hash)

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
using namespace std;

typedef unsigned long long ull;
const int E=10007;
const int LEN=1000002;
int len,Ans,l;
char c[LEN];
ull power[LEN],h[LEN];

int main()
{
	power[0]=1;
	for(int i=1;i<LEN;++i) power[i]=power[i-1]*E;
	for(scanf("%s",c+1);c[1]!='.';scanf("%s",c+1))
	{
		len=strlen(c+1);
		for(int i=1;i<=len;++i) h[i]=h[i-1]*E+c[i];
		Ans=1;
		for(int i=len;i>=2;--i)
		{
			if(len%i) continue;
			l=len/i;
			if(h[len-l]==(h[len]-h[l]*power[len-l]))
			{Ans=i; break;}
		}
		printf("%d
",Ans);
	}
	return 0;
}