Prime Ring Problem

Prime Ring Problem

A ring is compose of n circles as shown in diagram. Put natural number 1, 2, ..., n into each circle separately, and the sum of numbers in two adjacent circles should be a prime.

Note: the number of first circle should always be 1.


Inputn

(0 < n < 20).
Output

　　The output format is shown as sample below. Each row represents a series of circle numbers in the ring beginning from 1 clockwisely and anticlockwisely. The order of numbers must satisfy the above requirements. Print solutions in lexicographical order. 

You are to write a program that completes above process.

Print a blank line after each case.
Sample Input

6
8

Sample Output

Case 1:
1 4 3 2 5 6
1 6 5 2 3 4

Case 2:
1 2 3 8 5 6 7 4
1 2 5 8 3 4 7 6
1 4 7 6 5 8 3 2
1 6 7 4 3 8 5 2

 

#include <cstdio>
#include <cmath>
#include <cstring>
#define maxn 20
bool mark[maxn+5];
int key[maxn+5];
bool dp[maxn][maxn];
int n;
void is_prime()
{
    //memset(dp,true,sizeof(dp));
    for(int i = 1;i < 20; i++)
        for(int j = 1;j < 20;j++)
        {
            int cmp = i + j;
            int flag = 0;
            for(int k = 2; k <= sqrt(cmp); k++)
            {
                if(cmp % k == 0)
                {
                    dp[i][j] = false;
                    flag = 1;
                    break;
                }
            }
            if(!flag)
                dp[i][j] = true;
        }
}

void print()
{
    for(int i = 1;i <= n; i++)
    {
        if(i != 1)
            printf(" ");
        printf("%d",key[i]);
    }
    printf("\n");
}

void dfs(int ans)
{
    if(ans == n && dp[key[ans]][1])
    {
        print();
    }
    for(int i = 2; i <= n; i++)
    {
        if(mark[i] == false)
        {
            if(dp[i][key[ans]])
            {
                mark[i] = true;
                key[ans+1] = i;
                dfs(ans+1);
                mark[i] = false;
            }
        }
    }
}

int main()
{
    int count = 0;
    memset(dp,false,sizeof(dp));
    is_prime();
    while(scanf("%d",&n) != EOF)
    {
        printf("Case %d:\n",++count);
        memset(mark,false,sizeof(mark));
        key[1] = 1;
        dfs(1);
        printf("\n");
    }
    return 0;
}

 

#include <cstdio>
#include <cmath>
#include <cstring>
#define maxn 20
bool mark[maxn+5];
int key[maxn+5];
bool dp[maxn][maxn];
int n;
void is_prime()
{
    memset(dp,true,sizeof(dp));
    for(int i = 1;i < 20; i++)
        for(int j = 1;j < 20;j++)
        {
            int cmp = i + j;
            for(int k = 2; k <= sqrt(cmp); k++)
            {
                if(cmp % k == 0)
                {
                    dp[i][j] = false;
                    break;
                }
            }
        }
}

void print()
{
    for(int i = 1;i <= n; i++)
    {
        if(i != 1)
            printf(" ");
        printf("%d",key[i]);
    }
    printf("\n");
}

void dfs(int ans)
{
    if(ans == n && dp[key[ans]][1])
    {
        print();
    }
    for(int i = 2; i <= n; i++)
    {
        if(mark[i] == false)
        {
            if(dp[i][key[ans]])
            {
                mark[i] = true;
                key[ans+1] = i;
                dfs(ans+1);
                mark[i] = false;
            }
        }
    }
}

int main()
{
    int count = 0;
    is_prime();
    while(scanf("%d",&n) != EOF)
    {
        printf("Case %d:\n",++count);
        key[1] = 1;
        dfs(1);
        printf("\n");
    }
    return 0;
}