文章结束给大家来个程序员笑话：[M]
标题描述： 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. 输入： n (1 < n < 17). 输出： 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. 样例输入： 6 8 样例输出： 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 提示： 用printf打印输出。
#include <stdio.h> #include <stdlib.h> #include <string.h> int prime[35]; int visit[18]; int record[18]; void pre_process() { int i, j; memset(prime, 1, sizeof(prime)); prime[0] = prime[1] = 0; for (i = 2; i <= 34; i ++) { if (prime[i]) { for (j = i + i; j <= 34; j += i) { prime[j] = 0; } } } } void depth_first_search(int i, int n) { int j; if (i == n && prime[1 + record[n - 1]]) { // 构成素数环（do not forge test the first data and the last data） for (j = 0; j < n; j ++) { if (j == n - 1) printf("%d\n", record[j]); else printf("%d ", record[j]); } } else { // 深度优先遍历 for (j = 2; j <= n; j ++) { if (visit[j] == 0 && prime[record[i - 1] + j]) { // j没有访问过，并且和前一个记载之和为素数 record[i] = j; visit[j] = 1; depth_first_search(i + 1, n); visit[j] = 0; } } } } int main() { int i, n; i = 0; pre_process(); while (scanf("%d", &n) != EOF) { printf("Case %d:\n", ++ i); memset(visit, 0, sizeof(visit)); record[0] = 1; visit[1] = 1; depth_first_search(1, n); printf("\n"); } return 0; } /************************************************************** Problem: 1459 User: wangzhengyi Language: C Result: Accepted Time:410 ms Memory:912 kb ****************************************************************/
文章结束给大家分享下程序员的一些笑话语录：
一个合格的程序员是不会写出 诸如 “摧毁地球” 这样的程序的，他们会写一个函数叫 “摧毁行星”而把地球当一个参数传进去。