HDOJ_ACM_Prime Ring Problem

 

Problem Description

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.

 

Input

n (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

 

Codes

#include <stdio.h>
#include <string.h>
#define N 20
int primeMap[N + N];  //record which num is prime
int isUsed[N + 5];    //judge if the num is used 
int store[N + 5];     //record the path
int n, count = 1; 
void Initializing()  // 0 means prime, 1 means not prime
{
    int i, j;
    for (i = 4; i < 40; i++)
    {
        for (j = 2; j <= i / 2; j++)
        {
            if (i % j == 0)
            {
                primeMap[i] = 1;
                break;
            }
        }
    }
}
void DFS(int step)
{
    int i;
    //the exit condition
    if (step == n && primeMap[store[step] + 1] == 0)
    {
        for(i = 1; i < n; i++)
            printf("%d ",store[i]);
        printf("%d\n", store[n]);
    }
    //find the available num
    for (i = 2; i <= n; i++)
    {
        //is not prime or the num is used
        if (primeMap[store[step] + i] == 1 || isUsed[i] == 1)
            continue;
        //recursion
        isUsed[i] = 1;
        store[step + 1] = i; 
        DFS(step + 1);
        isUsed[i] = 0;
    }
}
int main()
{
    Initializing();
    while (scanf("%d",&n) != EOF)
    {
        memset(isUsed,0,sizeof(isUsed));
        store[1] = 1;
        printf("Case %d:\n", count++);
        //remove the odd, because it's useless
        if (n % 2 == 0)
            DFS(1);
        puts("");
    }
    return 0;
}

 

 

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