## sicily 1002 Anti-prime Sequences 原

Ciel

### 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``````

## 代码：

``````// Problem#: 1002
// Submission#: 1792032
// All Copyright reserved by Informatic Lab of Sun Yat-sen University
#include<iostream>
#include<cstring>
using namespace std;

#define MAX 10000
#define N 1010

bool isPrime[MAX];
int prime[MAX];
bool visit[N];
int re[N];
int n,m,d,t,sum;
bool flag;

void primeList(){
memset(isPrime,true,sizeof(isPrime));
for(int i = 2;i <= MAX;++i){
if(isPrime[i])  prime[++prime[0]] = i;
for(int j = 1,k;j <= MAX && (k = i * prime[j]) <= MAX;++j)
{
isPrime[k] = false;
if(i % prime[j] == 0)   break;
}
}
}

bool judge(int len){
int sum;
if(len <= 1)    return true;
for(int i = 0;i < len;++i){
sum = 0;
if(i <= len - d){
for(int j = i;j - i + 1<= d ;++j){
sum += re[j];
if(j-i+1 > 1 && isPrime[sum])   return false;
}
}else{
for(int j = i;j < len;++j){
sum += re[j];
if(j-i+1 > 1 && isPrime[sum])   return false;
}
}
}
return true;
}

void dfs(int num){
if(++t > 4000)  return;
if(flag)    return;
if(!judge(num)) return;
if(num== m - n + 1){
flag = 1;
return;
}
for(int i = n;i <= m && !flag;++i){
if(visit[i])    continue;
re[num] = i;
visit[i] = 1;
dfs(num+1);
visit[i] = 0;
}
}

int main(){
primeList();
while(cin>>n>>m>>d && n && m && d){
flag = false;
t = 0;
memset(visit,0,sizeof(visit));
dfs(0);
if(flag){
cout << re[0];
for( int i=1 ; i<m-n+1 ; ++i ) cout << "," << re[i];
cout << endl;
}else cout << "No anti-prime sequence exists." << endl;
}
return 0;
}``````

### Ciel

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