#include<iostream> #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> using namespace std; int n; int a[30]; int prime[]={0,0,1,1,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0,0,0}; int vis[30]; int dfs(int ans) { a[1]=1; if(ans==n+1&&prime[a[1]+a[ans-1]]){ for(int i=1;i<n;i++){ printf("%d ",a[i]); } printf("%d\n",a[n]); } else{ for(int i=2;i<=n;i++){ if(!vis[i]&&prime[i+a[ans-1]]){ a[ans]=i; vis[i]=1; dfs(ans+1); vis[i]=0; } } } } int main() { int k; while(scanf("%d",&n)!=EOF){ printf("Case %d:\n",k++); dfs(2); printf("\n"); } return 0; }
这个题的关键是在于判断条件和vis标记的所处位置。其他的就是一个不需要太变形的dfs题。