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Description
A ring is composed of n (even number) circles as shown in diagram. Put natural numbers 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.
You are to write a program that completes above process.
6
8
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<cstring>
#include<algorithm>
using namespace std;
int is_prime(int n)
{
for(int k= 2; k*k <= n; k++)
if(n% k== 0) return 0;
return 1;
}
int n, A[50], isp[50], vis[50];
void dfs(int cur)
{
if(cur == n && isp[A[0]+A[n-1]])
{
for(int i = 0; i < n; i++)
{
if(i != 0)
printf(" ");
printf("%d", A[i]);
}
printf("\n");
}
else
for(int i = 2; i <= n; i++)
if(!vis[i] && isp[i+A[cur-1]])
{
A[cur] = i;
vis[i] = 1;
dfs(cur+1);
vis[i] = 0;
}
}
int main()
{
int kase = 0;
while(scanf("%d", &n) == 1 && n > 0)
{
if(kase > 0)
printf("\n");
printf("Case %d:\n", ++kase);
for(int i = 2; i <= n*2; i++)
isp[i] = is_prime(i);
memset(vis, 0, sizeof(vis));
A[0] = 1;
dfs(1);
}
return 0;
}
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原文地址：http://www.cnblogs.com/xl1164191281/p/4696708.html
