HDU 1016 Prime Ring Problem 題解

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Problem 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 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

本題就是考遞迴搜尋的能力。

資料不大，其他Prime， map等的最佳化都沒多大作用的。

記得記錄好資料，就不會有問題了。

不過HDU的判斷系統的確垃圾，其他OJ都不會在意末尾多一個空格或者斷行符號的問題的，HDU就一個空格一個斷行符號都一定要按照她的格式，否則就presentation error.

#include <stdio.h>const int MAX_N = 20;int num;int cycle[MAX_N];bool vis[MAX_N];bool isPrime(int n){for (int i = 2; i*i <= n; i++)if (n % i == 0) return false;return true;}bool isLegal(int val, int i){int left = i-1;if (!isPrime(val+cycle[left])) return false;if (i+1 == num && !isPrime(val+cycle[0])) return false;return true;}void printNums(int i = 1){if (i == num){for (int j = 0; j+1 < num; j++){printf("%d ", cycle[j]);}printf("%d\n", cycle[i-1]);return ;}for (int v = 2; v <= num; v++){if (vis[v]) continue;if (isLegal(v, i)){cycle[i] = v;vis[v] = true;printNums(i+1);vis[v] = false;}}}int main(){int t = 0;cycle[0] = 1;while (scanf("%d", &num) != EOF){printf("Case %d:\n", ++t);for (int i = 2; i <= num; i++) vis[i] = false;printNums();putchar('\n');}return 0;}