A - Number Sequence

Given two sequences of numbers : a[1], a[2], …… , a[N], and b[1], b[2], …… , b[M] (1 <= M <= 10000, 1 <= N <= 1000000). Your task is to find a number K which make a[K] = b[1], a[K + 1] = b[2], …… , a[K + M - 1] = b[M]. If there are more than one K exist, output the smallest one.

Input

The first line of input is a number T which indicate the number of cases. Each case contains three lines. The first line is two numbers N and M (1 <= M <= 10000, 1 <= N <= 1000000). The second line contains N integers which indicate a[1], a[2], …… , a[N]. The third line contains M integers which indicate b[1], b[2], …… , b[M]. All integers are in the range of [-1000000, 1000000].

Output

For each test case, you should output one line which only contain K described above. If no such K exists, output -1 instead.

Sample Input

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7
2
13 5
1 2 1 2 3 1 2 3 1 3 2 1 2
1 2 3 1 3
13 5
1 2 1 2 3 1 2 3 1 3 2 1 2
1 2 3 2 1

Sample Output

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-1

题目大意：找到子串的位置，返回第一个下标，否则返回-1

参考代码

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#include <iostream>
#include <cstdio>
#include <algorithm>
#include <stdio.h>
#include <string>
#include <cstring>
using namespace std;
int ne[10005];
int t[10005];
int s[1000005];
int lens,lent;
void getne()
{
	int i,j;
	j=-1,i=0;
	ne[0]=-1;
	//int len=strlen(t);
	while(i<lent)
	{
		if(j==-1||t[i]==t[j])
		{
			j++;
			i++;
			ne[i]=j;
		}
		else
		{
			j=ne[j];
		}
	}
}
int KMP()
{
	int i,j;
	j=0,i=0;

	while(i<lens&&j<lent)
	{
		if(j==-1||s[i]==t[j]) 
		{
			i++;
			j++;
		}
		else
		{
			j=ne[j];
		}
	}
	if(j==lent)
	{
		return i-j+1;
	}
	else
	return -1;
}

int main()
{
	int n;
	int m,x;
	
	cin>>n;
	while(n--)
	{
		
		cin>>m>>x;
		for(int i=0;i<m;i++)
		{
			scanf("%d",&s[i]);
		
		}
		for(int i=0;i<x;i++)
		{
			scanf("%d",&t[i]);
		}
		memset(ne,0,sizeof(ne));
		lent=x;
		lens=m;
		getne();
		cout<<KMP()<<endl;
	}
	return 0;
	
}

B-Oulipo

The French author Georges Perec (1936–1982) once wrote a book, La disparition, without the letter ‘e’. He was a member of the Oulipo group. A quote from the book:

Tout avait Pair normal, mais tout s’affirmait faux. Tout avait Fair normal, d’abord, puis surgissait l’inhumain, l’affolant. Il aurait voulu savoir où s’articulait l’association qui l’unissait au roman : stir son tapis, assaillant à tout instant son imagination, l’intuition d’un tabou, la vision d’un mal obscur, d’un quoi vacant, d’un non-dit : la vision, l’avision d’un oubli commandant tout, où s’abolissait la raison : tout avait l’air normal mais…

Perec would probably have scored high (or rather, low) in the following contest. People are asked to write a perhaps even meaningful text on some subject with as few occurrences of a given “word” as possible. Our task is to provide the jury with a program that counts these occurrences, in order to obtain a ranking of the competitors. These competitors often write very long texts with nonsense meaning; a sequence of 500,000 consecutive ‘T’s is not unusual. And they never use spaces.

So we want to quickly find out how often a word, i.e., a given string, occurs in a text. More formally: given the alphabet {‘A’, ‘B’, ‘C’, …, ‘Z’} and two finite strings over that alphabet, a word W and a text T, count the number of occurrences of W in T. All the consecutive characters of W must exactly match consecutive characters of T. Occurrences may overlap.

Input

The first line of the input file contains a single number: the number of test cases to follow. Each test case has the following format:

One line with the word W, a string over {‘A’, ‘B’, ‘C’, …, ‘Z’}, with 1 ≤ |W| ≤ 10,000 (here |W| denotes the length of the string W).
One line with the text T, a string over {‘A’, ‘B’, ‘C’, …, ‘Z’}, with |W| ≤ |T| ≤ 1,000,000.

Output

For every test case in the input file, the output should contain a single number, on a single line: the number of occurrences of the word W in the text T.

Sample Input

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3
BAPC
BAPC
AZA
AZAZAZA
VERDI
AVERDXIVYERDIAN

Sample Output

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3
1
3
0

题目大意：求子串在母串中出现的次数

和上一题差不多，只需要统计即可

参考代码：

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#include <iostream>
#include <cstdio>
#include <algorithm>
#include <stdio.h>
#include <string>
#include <cstring>
using namespace std;
int ne[10005];
string s,t;
int lens,lent;
void getne()
{
	int i,j;
	j=-1,i=0;
	ne[0]=-1;
	while(i<lent)
	{
		if(j==-1||t[i]==t[j])
		{
			j++;
			i++;
			ne[i]=j;
		}
		else
		{
			j=ne[j];
		}
	}
}
int KMPCOUNT()
{
	int i,j;
	j=0,i=0;
	int count=0;
	if(lent==1&&lens==1)
	{
		if(s[0]==t[0])
		return 1;
		else
		return 0;
	}
	while(i<lens&&j<lent)
	{
		if(j==-1||s[i]==t[j]) 
		{
			i++;
			j++;
		}
		else
		{
			j=ne[j];
		}
		if(j==lent)
		{
			j=ne[j];
			count++;
		}
	}

	return count;
}

int main()
{
	int n;
	cin>>n;
	while(n--)
	{
		cin>>t;
		cin>>s;
		lent=t.size();
		lens=s.size();
		getne();
		cout<<KMPCOUNT()<<endl;
	}
	
}

C - 剪花布条

一块花布条，里面有些图案，另有一块直接可用的小饰条，里面也有一些图案。对于给定的花布条和小饰条，计算一下能从花布条中尽可能剪出几块小饰条来呢？

Input

输入中含有一些数据，分别是成对出现的花布条和小饰条，其布条都是用可见ASCII字符表示的，可见的ASCII字符有多少个，布条的花纹也有多少种花样。花纹条和小饰条不会超过1000个字符长。如果遇见#字符，则不再进行工作。

Output

输出能从花纹布中剪出的最多小饰条个数，如果一块都没有，那就老老实实输出0，每个结果之间应换行。

Sample Input

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abcde a3
aaaaaa  aa
#

Sample Output

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0
3

这题和上一题略有不同，因为是减花布条，母串剪完就没有了，所以如果剪出一个布条以后，j要回溯到-1位置。

参考代码：

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#include <iostream>
#include <cstdio>
#include <algorithm>
#include <stdio.h>
#include <string>
#include <cstring>
using namespace std;
int ne[10005];
string s,t;
int lens,lent;
void getne()
{
	int i,j;
	j=-1,i=0;
	ne[0]=-1;
	while(i<lent)
	{
		if(j==-1||t[i]==t[j])
		{
			j++;
			i++;
			ne[i]=j;
		}
		else
		{
			j=ne[j];
		}
	}
}
int KMPCOUNT()
{
	int i,j;
	j=0,i=0;
	int count=0;
	if(lent==1&&lens==1)
	{
		if(s[0]==t[0])
		return 1;
		else
		return 0;
	}
	while(i<lens&&j<lent)
	{
		if(j==-1||s[i]==t[j]) 
		{
			i++;
			j++;
		}
		else
		{
			j=ne[j];
		}
		if(j==lent)
		{
			j=-1;
			i--;//i在++以后已经不匹配了，要减1，回溯到上一步
			count++;
		}
	}

	return count;
}

int main()
{
	while(cin>>s>>t&&s[0]!='#')
	{
		lent=t.size();
		lens=s.size();
		getne();
		cout<<KMPCOUNT()<<endl;

	}
	
}

exkmp

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#include<string>
#include <iostream>

using namespace std;
string s,x;
int lenx,lens;
int ne[1005];
int ex[1005];
void getne (){	
	ne[0] = lenx;
	int a = 0 , p = 0;
	for (int i=1;i<lenx ; i++){
		if(i>=p || i+ne [i-a] >=p) {
			p= max(p,i);
			while(p<lenx && x[p]==x[p-i]) p++;
			ne[i] = p-i;
		a = i;
		}
		else ne [i] = ne[i-a];
	}
}
void getex() {// s的每个后缀与x匹配前缀 

	getne () ;
	int a = 0 , p = 0;
	for (int i=0;i<lens; i++){
		if(i>=p || i+ne [i-a] >=p) {
			p = max(p,i);
			while (p<lens &&p-i<lenx && s[p]==x[p-i]) p++;
			ex[i]= p-i;
			a = i;
		}
		else ex[i]= ne [i-a] ;
	}
}
int main()
{
	cin>>x>>s;
	lenx=x.size();
	lens=x.size();
	getne();
	getex();
	for(int i=0;i<8;i++)
	{
		cout<<ne[i]<<" "; 
	}
	cout<<endl;
	for(int i=0;i<6;i++)
	cout<<ex[i]<<" ";
	cout<<endl;
	
}
//x:aaaaac
//S:aaaaabbb

D-Power Strings

Given two strings a and b we define ab to be their concatenation. For example, if a = “abc” and b = “def” then ab = “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

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abcd
aaaa
ababab
.

Sample Output

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1
4
3

Hint

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

题目大意：求字符串中子串循环出现的最大次数

这题我最开始是把字符串切片，然后一个一个统计，如果大于之前的次数就更新次数，但是运行超时了，也不奇怪，1000000个字母，不超时才怪。

这题主要考察的是对next数组的理解。定义一个j=i-ne[i],这个就是之前匹配的最大前缀长度。用i%j,如果能整除，说明一定能输出

参考代码：

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#include<string>
#include <iostream>
#include <cstdio>
#include<cctype>
#include <string>
#include<string.h>
using namespace std;
string s,t;
int ne[100005];
int len;
void getne()
{
	int i,j;
	j=-1,i=0;
	ne[0]=-1;
	while(i<len)
	{
		if(j==-1||s[i]==s[j])
		{
			j++;
			i++;
			ne[i]=j;
		}
		else
		{
			j=ne[j];
		}
	}
}

int main(){
	while(cin>>s&&s!=".")
	{
		memset(ne,-1,sizeof(ne));
		len=s.size();
		getne();
		if(len%(len-ne[len])==0)
            printf("%d\n",len/(len-ne[len]));
        else
           printf("1\n");
	}
	
	
	return 0;
}