1016 Prime Ring Problem

最新推荐文章于 2021-08-13 17:41:59 发布

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本文介绍了一种利用深度优先搜索解决素数环问题的方法。该问题要求将1至n的数字填入由n个圆圈构成的环中，使得相邻两圆圈内的数字之和均为素数，并且整个环从数字1开始。通过预设素数数组来优化判断过程，确保了高效求解。

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：使用深度优先搜索

注意：1、素数的定义，2也是素数；此外最后一个数除了与前面的数的和为素数外，还需于第一个数的和为素数，因为是环；

技巧：数组isPrime[n] 表示n是否为素数，1为是0为否，这样直接操作数组下标就可以判断该数是否为素数，eg：isPrime[result[curr-1]+i]==1