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?'\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:\n",++cnt);
DFS(2);
printf("\n");
}
}

#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?'\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:\n",++cnt);

        DFS(2);

        printf("\n");

    }

}

巴特西

搜索专题: HDU1016Prime Ring Problem

Prime Ring Problem

最新文章

热门文章