题目链接:Uva 552

思路分析:时间限制为3s,数据较小,使用深度搜索查找所有的解。

代码如下:

#include <iostream>
#include <string.h>
using namespace std; const int MAX_N = ;
int n;
int A[MAX_N], vis[MAX_N]; int is_prime( int n )
{
if ( n == )
return true; for ( int i = ; i * i <= n; ++i )
if ( n % i == )
return false; return true;
} void dfs( int cur )
{
if ( cur == n && is_prime( A[] + A[n - ] ) )
{
int i = ; for ( i = ; i < n - ; ++i )
printf( "%d ", A[i] );
printf( "%d", A[i] );
printf( "\n" );
}
else
{
for ( int i = ; i <= n; ++i )
{
if ( !vis[i] && is_prime( i + A[cur - ] ) )
{
A[cur] = i;
vis[i] = ;
dfs( cur + );
vis[i] = ;
}
}
}
} int main( )
{
int count = ; while ( scanf( "%d", &n ) != EOF )
{
if ( count != )
printf( "\n" ); A[] = ;
memset( vis, , sizeof( vis ) );
printf( "Case %d:\n", count++ );
dfs( );
} return ;
}

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