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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<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<string>
#include<algorithm>
#include<queue>
using namespace std;
int n;
int a[25];//保存结果
int flag[25];//记录一个数是否已用过
bool isprime(int n)
{
if (n == 1 || n == 0)
return 0;
if (n == 2)
return 1;
for (int i = 2; i <= sqrt((double)n); i++)
{
if (n%i == 0)
return 0;
}
return 1;
}
void dfs(int tot)
{
if (tot == n-1)
{
if (isprime(a[n - 1] + a[0]))
{
int o = 0;
for (int i = 0; i < n; i++)
{
while (o++)
{
printf(" ");
break;
}
printf("%d", a[i]);
}
printf("\n");
}
return;
}
for (int i = 2; i <= n; i++)
{
if (flag[i] == 0)
{
if (isprime(a[tot] + i))
{
flag[i] = 1;
a[tot + 1] = i;
dfs(tot + 1);
flag[i] = 0;
}
}
}
}
int main()
{
int k = 1;
while (~scanf("%d", &n))
{
printf("Case %d:\n", k++);
a[0] = 1;
memset(flag, 0, sizeof(flag));
dfs(0);
printf("\n");
}
return 0;
}Prime Ring Problem—hdu1016(DFS)
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原文地址：http://blog.csdn.net/jelly_acm/article/details/51331381
