1016 Prime Ring Problem

Prime Ring Problem

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

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

 

#include <stdio.h> #include <string.h> #define MAXLEN 40 #define DATALEN 20 int isPrime[MAXLEN]; int visited[DATALEN]; int result[DATALEN]; void DFS(int curr, int num); int main(void) { int n; int count = 0; memset(isPrime, 0, MAXLEN * sizeof(int)); isPrime[3]=isPrime[5]=isPrime[7]=isPrime[11]=isPrime[13]=isPrime[17]=isPrime[19]=isPrime[23]=isPrime[29]=isPrime[31]=isPrime[37]=1; while(scanf("%d", &n) == 1) { count++; memset(visited, 0, DATALEN * sizeof(int)); memset(result, 0, DATALEN * sizeof(int)); visited[1] = 1; result[1] = 1; printf("Case %d:/n", count); DFS(2, n); printf("/n"); } } void DFS(int curr, int num) { int i, j; for(i = 2; i <= num; i++) { if(visited[i]==0 && isPrime[result[curr-1]+i]==1) { visited[i] = 1; result[curr] = i; if(curr != num) { DFS(curr+1, num); } else if(isPrime[i+1]==1) { for(j = 1; j < num; j++) { printf("%d ", result[j]); } printf("%d/n", result[j]); } visited[i] = 0; result[curr] = 0; } } } 

 

 

 

方法：使用深度優先搜索

注意：1、素數的定義，2也是素數；此外最後一個數除了與前面的數的和爲素數外，還需於第一個數的和爲素數，因爲是環；

技巧：數組isPrime[n] 表示n是否爲素數，1爲是0爲否，這樣直接操作數組下標就可以判斷該數是否爲素數，eg：isPrime[result[curr-1]+i]==1