\(\color{CornflowerBlue}{Problem \; Description}\)
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 \(1\).
\(\color{CornflowerBlue}{Input}\)
\(n (0 < n < 20).\)
\(\color{CornflowerBlue}{Output}\)
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.
\(\color{CornflowerBlue}{Sample\;Input}\)
6
8
\(\color{CornflowerBlue}{Sample\;Output}\)
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
\(\color{CornflowerBlue}{Source}\)
Asia 1996, Shanghai (Mainland China)
\(\color{CornflowerBlue}{Recommend}\)
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题目描述
如图，给定一个大小为\(n\)的圆环，将正整数\(1,2,\cdots,n\)放入圆环中，使得相邻圆环上的两数相加总为质数
注意：第一个圆环上的数字总为\(1\)
输入
\(n (0 < n < 20).\)
输出
按照字典序，输出所有可能的方案；具体输出格式见样例；
Pz's solution
1.首先，考虑到最大能拼成的数为\(18+19=37\)，可以先预处理出\(1 \sim 40\)之间的质数；
2.在保证前一项与当前搜索项相加为质数的情况下进行搜索；
3.当搜索到最后一个数时，特判一下\(1\)与之相加是否为质数即可；
- TAG：搜索；质数
PZ.cpp
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define N 40
int n,T,ans[N],p[N];
bool vis[N],used[N];
void init(){
vis[0]=vis[1]=1;
for(int i=2;i<N;++i){
if(!vis[i]) p[++p[0]]=i;
for(int j=1;j<=p[0]&&p[j]*i<N;++j){
vis[p[j]*i]=1;
if(i%p[j]==0) break;
}
}
}
void dfs(int res){
if(res==n){
if(!vis[1+ans[n]])
for(int i=1;i<=n;++i){
printf("%d%c",ans[i],(i==n ? '\n' : ' '));
}
return;
}
for(int i=2;i<=n;++i)
if(!used[i])
if(!vis[ans[res]+i]){
ans[res+1]=i;
used[i]=1;
dfs(res+1);
used[i]=0;
}
}
int main(){
init();
while(scanf("%d",&n)!=EOF){
++T;
memset(used,0,sizeof(used));
memset(ans,0,sizeof(ans));
ans[1]=1;
printf("Case %d:\n",T);
dfs(1);
putchar('\n');//注意样例输出格式
}
return 0;
}