标签：write series RoCE net graphic div number cal and
Inputn (0 < n < 20).
OutputThe output format is shown as sample below. Each row represents a series of circle numbers in the ring beginning from 1 clockwisely and anticlockwisely. The order of numbers must satisfy the above requirements. Print solutions in lexicographical order.
You are to write a program that completes above process.
Print a blank line after each case.
Sample Input
6 8
Sample Output
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
1 #include<iostream> 2 #include<stdio.h> 3 #include<cstring> 4 #include<queue> 5 6 using namespace std; 7 8 9 int n; 10 int a[23]; 11 int vis[23]; 12 bool mark[1000]; 13 14 void init() // 素数筛 15 { 16 for(int i = 2; i < 1000; ++i) 17 mark[i] = true; 18 19 for(int i = 2; i < 1000; ++i) 20 { 21 if(mark[i] == true) 22 { 23 for(int j = i*i; j < 1000; j += i) 24 { 25 mark[j] = false; 26 } 27 } 28 } 29 } 30 31 // 边枚举边判断，不要最后一次性判断，会超时 32 void dfs(int step) 33 { 34 if(step > 2) 35 { 36 if(mark[a[step-1]+a[step-2]] == false) // 判断最后两个数 37 return; 38 } 39 40 if(step == n+1) 41 { 42 if(mark[a[n]+a[1]] == false) // 判断最后一个数与第一个数 43 return; 44 for(int i = 1; i < n; ++i) 45 printf("%d ", a[i]); 46 printf("%d\n", a[n]); 47 48 } 49 50 for(int i = 2; i <= n; ++i) 51 { 52 if(!vis[i]) 53 { 54 a[step] = i; 55 vis[i] = 1; 56 dfs(step+1); 57 vis[i] = 0; 58 } 59 60 } 61 } 62 63 64 int main() 65 { 66 init(); 67 int cas = 1; 68 a[1] = 1; 69 while(scanf("%d", &n) != EOF) 70 { 71 printf("Case %d:\n", cas++); 72 memset(vis, 0, sizeof(vis)); 73 dfs(2); 74 printf("\n"); 75 } 76 77 78 return 0; 79 }
Prime Ring Problem HDU - 1016 (dfs)
标签：write series RoCE net graphic div number cal and
原文地址：https://www.cnblogs.com/FengZeng666/p/11517509.html
