搜索专题: HDU1016Prime Ring Problem

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 52146 Accepted Submission(s): 23096

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

Source

Asia 1996, Shanghai (Mainland China)

Problem : 1016 ( Prime Ring Problem ) Judge Status : Accepted
RunId : 21243288 Language : G++ Author : hnustwanghe
Code Render Status : Rendered By HDOJ G++ Code Render Version 0.01 Beta

#include<iostream>
#include<cstring>
#include<cstdio>

using namespace std;
const int N = 20 + 5;
bool prime[N*3],used[N]={0};
int a[N],n;
void prime_factor(){
    memset(prime,0,sizeof(prime));
    prime[0] = prime[1] = 1;
    for(int i=2;i<=N*2;i++)
    if(!prime[i]){
        for(int j=i*i;j<=N;j+=i)
            prime[j] = 1;
    }
}

void DFS(int cur){
    if(cur > n){
        if(!prime[a[1]+a[n]])
        for(int i=1;i<=n;i++)
            printf("%d%c",a[i],i==n?'
':' ');
        return;
    }
    for(int i=2;i<=n;i++){
            if(!used[i]){
                if(!prime[a[cur-1]+i]){
                    used[i] = true;
                    a[cur] = i;
                    DFS(cur+1);
                    used[i] = false;
                }
            }
    }
}

int main(){
    a[1] = 1;
    prime_factor();
    int cnt = 0;
    while(scanf("%d",&n)==1){
        printf("Case %d:
",++cnt);
        DFS(2);
        printf("
");
    }
}