POJ 2034 Anti-prime Sequences

链接: http://poj.org/problem?id=2034

Anti-prime Sequences

Time Limit: 3000MS Memory Limit: 30000K

Total Submissions: 3803 Accepted: 1726

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.

SampleInput

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

SampleOutput

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

Source

East Central North America 2004

Analyze

有数列 n, n+1, ..., m 还有k， 重新找一个数列，使得这个数列中的k个连续数字的和是合数。

这个就是裸DFS

Code

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

int n, m, k,sum;
int ans[1009];
bool is_prime[10009], vis[1009];
void init()
{
//因为题目要求是最多有10个小于1000
//所以你的素数最好初始化到10009，
//不然就会超时，嘿嘿。我在这里错咯几次
memset(is_prime, true, sizeof(is_prime));
is_prime[0] = is_prime[1] = false;
for(int i=2; i<10009; i++)
for(int j=2; i*j<10009; j++)
is_prime[i*j] = false;
}
bool place(int cur, int bak)
{
if(cur == 0)
return true;
int tmp = cur-k+1;
if(tmp < 0)
tmp = 0;
int sum = bak;
for(int i=cur-1; i>=tmp; i--)
{
sum += ans[i];
if(is_prime[sum])
return false;
}
return true;
}
void DFS(int cur)
{
if(sum)
return;
if(cur == m-n+1)
{
sum ++;
for(int i=0; i<m-n+1; i++)
printf("%d%c", ans[i], (i==(m-n)?'\n':','));
return;
}
for(int i=n; i<=m; i++)
{
if(!vis[i] && place(cur, i) )
{
ans[cur] = i;
vis[i] = true;
DFS(cur+1);
vis[i] = false;
}
}
return;
}
int main()
{
init();
while(scanf("%d%d%d", &n,&m,&k)!=EOF&&n&&m&&k)
{
memset(vis, false, sizeof(vis));
sum = 0;
DFS(0);
if(sum == 0)
printf("No anti-prime sequence exists.\n");
}
return 0;
}