标签：nat shanghai class names order ... format targe 训练
1、链接：
http://acm.hrbust.edu.cn/vj/index.php?c=problem-problem&id=2178
题目：
Description
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.
Input
n (0 < n < 20).
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. Print solutions in lexicographical order.
You are to write a program that completes above process.
Print a blank line after each case.
Sample Input
6
8
Sample Output
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
Source
Asia 1996, Shanghai (Mainland China)
2、解题分析:
题意：给定数n，要求1~n的n个数 从1开始 组成一个环，没相邻的两个数的和为质数（素数）。ps：首尾相接处也要判断
解法：利用dfs实现全排列，从全排列中筛选符合条件的，但是要一边DFS一边判断目前状态是否合法，可以实现优化。
3、代码
#include<bits/stdc++.h> using namespace std; int n; int step; int vis[105]; int ans[105]; int prime[12] = {2,3,5,7,11,13,17,19,23,29,31,37};//打表 bool primer_num(int x) //素数判断 { for(int i = 0; i < 12; i++) if(x == prime[i]) return true; return false; } void dfs(int step) { if(step == n + 1) { if(primer_num(ans[n] + ans[1])){ for(int i = 1; i <= n; i++) { printf("%d",ans[i]); if(i != n) printf(" "); else { printf("\n"); } } } } for(int i = 2 ; i <= n; i++) { if(vis[i] == 0 && primer_num(ans[step-1] + i)) { vis[i] = 1; ans[step] = i; dfs(step + 1); vis[i] = 0; } } } int main() { int cas = 1; while(~scanf("%d",&n)) { memset(vis, 0 ,sizeof(vis)); printf("Case %d:\n",cas++); ans[1] = 1; dfs(2); puts(""); } return 0; }
标签：nat shanghai class names order ... format targe 训练
原文地址：http://www.cnblogs.com/hhkobeww/p/7755856.html
