Prime Ring Problem HDU - 1016
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
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 思路:DFS遍历所有结果
AC Code:
#include<iostream>
#include<cstring>
using namespace std;
int num[];
int vis[];
int n;
bool isPrime (int k) {
for (int i = ; i * i <= k; i++)
if (k % i == )
return false;
return true;
}
void dfs (int k) {
if (k > n) {
if (isPrime (num[] + num[n])) {
for (int i = ; i < n; i++)
printf("%d ", num[i]);
printf("%d\n", num[n]);
}
return ;
}
for (int i = ; i <= n; i++) {
if (vis[i] == false) {
if (isPrime (num[k - ] + i)) {
num[k] = i;
vis[i] = true;
dfs (k + );
vis[i] = false;
}
}
}
return ;
}
int main() {
int sum = ;
while(~scanf ("%d", &n)){
printf ("Case %d:\n", sum++);
memset (vis, false, sizeof(vis));
num[] = ;
vis[] = true;
dfs();
putchar('\n');
}
}
Prime Ring Problem HDU - 1016的更多相关文章
- Prime Ring Problem HDU - 1016 (dfs)
Prime Ring Problem HDU - 1016 A ring is compose of n circles as shown in diagram. Put natural number ...
- HDU 1016 Prime Ring Problem(经典DFS+回溯)
Prime Ring Problem Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Other ...
- [HDU 1016]--Prime Ring Problem(回溯)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1016 Prime Ring Problem Time Limit: 4000/2000 MS (Jav ...
- HDU 1016 Prime Ring Problem(素数环问题)
传送门: http://acm.hdu.edu.cn/showproblem.php?pid=1016 Prime Ring Problem Time Limit: 4000/2000 MS (Jav ...
- HDOJ(HDU).1016 Prime Ring Problem (DFS)
HDOJ(HDU).1016 Prime Ring Problem (DFS) [从零开始DFS(3)] 从零开始DFS HDOJ.1342 Lotto [从零开始DFS(0)] - DFS思想与框架 ...
- hdu 1016 Prime Ring Problem(DFS)
Prime Ring Problem Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Other ...
- hdu 1016 Prime Ring Problem(深度优先搜索)
Prime Ring Problem Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Other ...
- HDU 1016 Prime Ring Problem (DFS)
Prime Ring Problem Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Other ...
- HDU 1016 Prime Ring Problem (回溯法)
Prime Ring Problem Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Other ...
随机推荐
- 终于明白vim 和 grep 中 的正则表达式的用法, vim 正则表达式 和grep基本正则表达式 几乎一样
要搞清楚 vim中的正则和普通的Perl正则表达式的区别: 因为在perl中所有的元字符 都可以直接使用, 不需要在 元字符的前面加 反斜杠. 但是在vim, 包括grep中就有所区别, 同样是元字符 ...
- SDOI2017相关分析 线段树
题目 https://loj.ac/problem/2005 思路 \[ \sum_{L}^{R}{(x_i-x)^{2}} \] \[ \sum_{L}^{R}{(x_i^2-2*x_i*x+x^{ ...
- luogu1975 [国家集训队]排队
思路 序列中 |i | 1| 2| 3| 4| 5| 6| 7| 8| 9| 10| |----|--|--|--|--|--|--|--|--|--|--| |a[i]| a| b| c| L| d ...
- HDU 4366 Successor(dfs序 + 分块)题解
题意:每个人都有一个上司,每个人都有能力值和忠诚值,0是老板,现在给出m个询问,每次询问给出一个x,要求你找到x的所有直系和非直系下属中能力比他高的最忠诚的人是谁 思路:因为树上查询很麻烦,所以我们直 ...
- (转)renren-fast解读(一)
(二期)8.renren-fast项目解读(一) [课程八]预防xss...注入.xmind0.2MB [课程八预习]开...解读.xmind0.5MB 课程八_日志处理与...模块.xmind0.2 ...
- dll注入到指定进程
talk is cheap,show me code 代码有详细注释 主程序 #include "stdafx.h" #include <windows.h> #inc ...
- Neo4j 文档
Graph Fundamentals 基础 Basic concepts to get you going. A graph database can store any kind of data u ...
- eclipse创建springBoot项目
创建Spring Boot 工程 先在eclipse中安装spring -tool -suite插件,然后根据以下步骤可以创建1.新建Spring Starter Project 2.Packagin ...
- POJ 3278 Catch That Cow(赶牛行动)
POJ 3278 Catch That Cow(赶牛行动) Time Limit: 1000MS Memory Limit: 65536K Description - 题目描述 Farmer J ...
- 简单实现RN调用原生方法(IOS)
在React Native中,一个“原生模块”就是一个实现了“RCTBridgeModule”协议的Objective-C类(个人理解RCTBridgeModule就是react与native之间的桥 ...