1 solutions

  • 0
    @ 2023-3-5 14:02:01
    #include<stdio.h>
    #include<string.h>
    const int N = 10 + 5;
    int count, n, vis[N];
    
    void dfs(int p, int m) {
    	if (m == n){
    		count++;
    		return ;
    	}
    	vis[p] = 2;
    	for(int i=1;i<=9;i++){
    		if (vis[i] == 1){
    			if ((vis[(p + i) / 2] == 0 || vis[(p + i) / 2] == 1) && (p + i == 10 || ((p == 1 && i == 7) || (p == 7 && i == 1)) || ((p == 1 && i == 3) || (p == 3 && i == 1)) || ((p == 3 && i == 9) || (p == 9 && i == 3)) || ((p == 7 && i == 9) || (p == 9 && i == 7))))
    				continue;
    			else
    				dfs(i, m + 1);
    		}
    	}
    	vis[p] = 1;
    }
    int main()
    {
    	scanf("%d",&n);
    	for(int i=1;i<=n;i++){
    		int t;
    		scanf("%d",&t);
    		vis[t]=1;
    	}
    	for(int i=1;i<=9;i++){
    		if(vis[i]==1){
    			dfs(i,1);
    		}
    	}
    	if(!count)printf("加油华为加油china\n");
    	else printf("%d\n",count);
    	return 0;
    }
    
    • 1

    Information

    ID
    6702
    Time
    1000ms
    Memory
    256MiB
    Difficulty
    9
    Tags
    (None)
    # Submissions
    7
    Accepted
    5
    Uploaded By