Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 37427 Accepted Submission(s): 16521
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
1.自然数（natural number），是非负整数（1, 2, 3, 4……）。认为自然数不包含零的其中一个理由是因为人们在开始学习数字的时候是由“一、二、三...”开始，而不是由“零、一、二、三...”开始, 因为这样是非常不自然的。在全球范围内，目前针对0是否属于自然数的争论依旧存在。在中国大陆，2000年左右之前的中小学教材一般不将0列入自然数之内，或称其属于“扩大的自然数列”。在2000年左右之后的新版中小学教材中，普遍将0列入自然数。
2.代码：
#include<cstdio>
#include<cstring>
using namespace std;
int n;
int vis[30];
int a[30];
int p[20]={2,3,5,7,11,13,17,19,23,29,31,37,41};
int c=1;
int cnt;
bool isPrime(int x)
{
for(int i=0;i<13;i++)
{
if(x==p[i])
{
return 1;
}
}
return 0;
}
void dfs(int level,int num)
{
a[level]=num;
vis[num]=1;
if(level==n)
{
if(isPrime(a[n]+a[1]))
{
if(cnt==0)
printf("Case %d:
",c++);
cnt++;
for(int i=1;i<=n;i++)
{
if(i==1)
printf("%d",a[i]);
else
printf(" %d",a[i]);
}
printf("
");
}
return;
}
for(int i=1;i<=n;i++)
{
if(vis[i]==1||isPrime(i+a[level])==0)
continue;
dfs(++level,i);
level--;
vis[i]=0;
}
}
int main()
{
while(scanf("%d",&n)==1)
{
memset(vis,0,sizeof(vis));
cnt=0;
dfs(1,1);
printf("
");
}
return 0;
}