uva 524 - Prime Ring Problem

A ring is composed of n (even number) circles as shown in diagram. Put natural numbers 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 <= 16)

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.

You are to write a program that completes above process.

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

#include <math.h>#include <stdio.h>int N;int result[50];int visit[50];int is_prime(int n){int max = sqrt(n);int i = 2;while(i <= max){if(n % i == 0){return 0;}i++;}return 1;}void print_result(void){int i = 1;while(i <= N){printf("%d ", result[i]);i++;}printf("\n");}void search(int current_count){if(current_count == N && is_prime(1 + result[current_count])){print_result();visit[result[current_count]] = 0;return;}int i;for(i = 2; i <= N; i++){if(visit[i] == 0  && is_prime(i + result[current_count]) ){visit[i] = 1;result[current_count + 1] = i;search(current_count + 1);}}visit[result[current_count]] = 0;}int main(int argc, char **argv){int i = 1;while(scanf("%d", &N) != EOF){result[1] = 1;visit[1] = 1;printf("Case %d:\n", i);i++;search(1);}return 0;}

这个代码可以从下面两点优化：

１．　预先把数字是否为质数保存到数组里面，而不是一个个的去计算。

２．　考虑到数字的奇偶行，下面一行可：

for(i = 2; i <= N; i++)

可以改为：

 for(i = (current_count+1) %2 + 2; i <= N; i+=2)

Ｒｅｆｅｒｅｎｃｅs:

1. http://www.programering.com/a/MTMwIDNwATU.html

2. http://www.tuicool.com/articles/Ezuum2a