[sicily online]1002. Anti-prime Sequences

Constraints

Time Limit: 3 secs, Memory Limit: 32 MB

Description

Given a sequence of consecutive integers n,n+1,n+2,...,m, an anti-prime sequence is a rearrangement of these integers so that each adjacent pair of integers sums to a composite (non-prime) number. For example, if n = 1 and m = 10, one such anti-prime sequence is 1,3,5,4,2,6,9,7,8,10. This is also the lexicographically first such sequence. We can extend the definition by defining a degree danti-prime sequence as one where all consecutive subsequences of length 2,3,...,d sum to a composite number. The sequence above is a degree 2 anti-prime sequence, but not a degree 3, since the subsequence 5, 4, 2 sums to 11. The lexicographically .rst degree 3 anti-prime sequence for these numbers is 1,3,5,4,6,2,10,8,7,9.

Input

Input will consist of multiple input sets. Each set will consist of three integers, n, m, and d on a single line. The values of n, m and d will satisfy 1 <= n < m <= 1000, and 2 <= d <= 10. The line 0 0 0 will indicate end of input and should not be processed.

Output

For each input set, output a single line consisting of a comma-separated list of integers forming a degree danti-prime sequence (do not insert any spaces and do not split the output over multiple lines). In the case where more than one anti-prime sequence exists, print the lexicographically first one (i.e., output the one with the lowest first value; in case of a tie, the lowest second value, etc.). In the case where no anti-prime sequence exists, output No anti-prime sequence exists.

Sample Input

1 10 2
1 10 3
1 10 5
40 60 7
0 0 0

Sample Output

1,3,5,4,2,6,9,7,8,10
1,3,5,4,6,2,10,8,7,9
No anti-prime sequence exists.
40,41,43,42,44,46,45,47,48,50,55,53,52,60,56,49,51,59,58,57,54

题目分析：刚看到这个题目时，就想按照排列算法，把所有情况遍历一遍，找到满足条件就停止遍历。但是超时超的严重，特别是在没有序列的情况下。优化了一天还是超时，最后求助于网上的大神http://blog.youkuaiyun.com/ChinaCzy/article/details/5673174

原来这个题目是用深度优先搜索做，仔细想了想，的确完美。里面有一个“卡时”的概念，虽然没有证明出来本题利用卡时的正确，但还是结果还是正确了。下面的代码是利用人家的思想，默写了一遍，应该算是大神的代码：

#include<iostream>
#include<fstream>
#include<vector>
#include<string>
#include<algorithm>
#include<cmath>
#include<string.h>
using namespace std;

const long size=10000;

bool com[size];
bool used[1010];
int x[1010];
bool ok;
int n,m,d,t;
void initComposite()
{//质数是false
	int tmp=sqrt((double)size);
	for(long i=2;i<=tmp;i++)
	{
		if(com[i]==true)continue;
		for(long j=i;i*j<size;j++)
			com[i*j]=true;
	}
}

void print()
{
	for(int i=0;i<m-n;i++)
		cout<<x[i]<<',';
	cout<<x[m-n]<<endl;
}

bool check(int deep)
{
	if(deep<2) return true;
	int tmpD=2;
	int i,flag=0;
	while(tmpD<=d)
	{
		for(i=0;i<=deep-tmpD;i++)
		{
			int sum=0;
			vector<int>::size_type j;
			for(j=0;j<tmpD;j++)
			{
				sum+=x[i+j];
			}//end for

			if(!com[sum])
			{
				flag=1;
				break;
			}//end if
		}//end for
		tmpD++;
		if(flag==1)
			return false;
	}
	return true;
}

void dfs(int deep)
{
	if(++t>4000)return ;
	if(ok)return ;
	if(!check(deep))return ;
	if(deep==m-n+1)
	{
		ok=true;
		return;
	}
	for(int i=n;i<=m&&!ok;++i)
	{
		if(used[i]) continue;
		used[i]=true;
		x[deep]=i;
		dfs(deep+1);
		used[i]=false;
	}//end for
}

int main()
{
	initComposite();
	while(cin>>n>>m>>d && n!=0)
	{
		ok=false;
		t=0;
		memset(used,0,sizeof(used));
		memset(x,0,sizeof(x));
		dfs(0);
		if(ok)print();
		else cout<<"No anti-prime sequence exists."<<endl;
	}//end while
}