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Prime Ring ProblemTime Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 22702 Accepted Submission(s): 10108Problem DescriptionA 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. Inputn (0 < n < 20). OutputThe 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 Input68 Sample OutputCase 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<iostream> #define MAX 25 using namespace std; int a[MAX], b[MAX], n, t; bool visit[MAX], prime[MAX*2]; int gcd(int x, int y){ if (y == 0)return x; return gcd(y, x%y); } void dfs(int i, int pos){ if (pos == n && prime[b[pos]+1]){ for (int i = 1; i < pos; i++){ cout << b[i] << " "; } cout << b[pos] << endl; return; } for (int j = 2; j <= n; j++){ if (!visit[j] && prime[b[i] + a[j]]){ visit[j] = true; b[i + 1] = a[j]; dfs(i + 1, pos + 1); visit[j] = false; } } return; } int main() { t = 0; while (cin >> n){ for (int i = 1; i <= n; i++){ a[i] = i; } fill(visit, visit + MAX, false); fill(prime, prime + MAX * 2, true); for (int i = 2; i < MAX * 2; i++){ if (prime[i]){ for (int j = i * 2; j < MAX * 2; j += i){ prime[j] = false; } } } cout << "Case " << ++t << ":" << endl; b[1] = 1; visit[1] = true; dfs(1, 1); cout << endl; } return 0; }
原文：http://www.cnblogs.com/littlehoom/p/3555216.html