Hduoj1016【深搜】【水题】

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 33730    Accepted Submission(s): 14932

Problem Description

A ring is compose of n circles as shown in diagram. Put natural number 1, 2, ..., n into each circle separately, and the sum of numbers in two adjacent circles should be a prime.

Note: the number of first circle should always be 1.

 

Input

n (0 < n < 20).

 

Output

The 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

68

 

Sample Output

Case 1:1 4 3 2 5 61 6 5 2 3 4Case 2:1 2 3 8 5 6 7 41 2 5 8 3 4 7 61 4 7 6 5 8 3 21 6 7 4 3 8 5 2

 

Source

Asia 1996, Shanghai (Mainland China) 

#include<stdio.h>#include<string.h>int prim[42], r[22], num, vis[22], n;void getprim(){prim[1] = 1;for(int i = 2; i < 8; ++i){for(int j = 2*i; j < 42; j += i)prim[j] = 1;}} void dfs(int x){if(num == n){if(!prim[r[num] + r[1]]){for(int i = 1; i < n; ++i)printf("%d ", r[i]);printf("%d\n", r[n]); }return ;}for(int i = 2; i <= n; ++i){if(!vis[i] && ! prim[r[num] + i]){r[++num] = i;vis[i] = 1;dfs(i);num--;vis[i] = 0;}}}int main(){int i, j, k, cas =  1;getprim();while(scanf("%d", &n) != EOF){memset(vis, 0, sizeof(vis));memset(r, 0, sizeof(r));r[1] = 1;num = 1;printf("Case %d:\n", cas++);dfs(1);printf("\n");}return 0;}

题意：给出一个数n，要求1~n，这n个数字构成一个素数环，即环中的任意前后2项之和为素数，输出这个环以1为开头的顺时针序列。

思路：先打素数表，再深搜。