- 题目描述：
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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：
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n (1 < n < 17).
- 输出：
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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：
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68
- 样例输出：
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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
为了解决该问题，我们可以采用回溯法枚举每一个值。当第一个数位为1确定时，我们尝试放入第二个数，使其和1的和为素数，放入后再尝试放入第三个数，使其与第二个数的和为素数，直到所有的数全部被放入环中，且最后一个数与1的和也是素数，那么这个方案即为答案，输出；若在尝试放数的过程中， 发现当前位置无论放置任何之前未被使用的数均不可能满足条件，那么我们回溯 改变其上一个数，直到产生我们所需要的答案，或者确实不再存在更多的解。
View Code#include "stdafx.h"#include
using namespace std;int prime[] = { 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37 };int number[20];bool hash[20];int n;bool isPrime(int num){ for (int i = 0; i < 12;i++) if (num == prime[i]) return true; return false;}void printArray(){ if (isPrime(number[n] + number[1]) == false) return; for (int i = 1; i <= n; i++) { if (i != 1) printf(" "); printf("%d", number[i]); } printf("\n");}void DFS(int num){ if (num > 1 && isPrime(number[num] + number[num - 1]) == false) return; if (num == n) { printArray(); return; } for (int i = 2; i <= n; i++) { if (hash[i] == false) { hash[i] = true; number[num + 1] = i; DFS(num + 1); hash[i] = false; } }}int main(){ int cas = 0; while (scanf("%d", &n) != EOF) { cas++; for (int i = 0; i < 20; i++) hash[i] = false; number[1] = 1; printf("Case %d:\n", cas); hash[1] = true; DFS(1); printf("\n"); } return 0;}