Prime Ring Problem
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 34318 Accepted Submission(s): 15175
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.
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)
Recommend
JGShining
题目分析：
就是输出两想加都等于素数的环
第一个代码是全排列的一个代码模板
#include <cstdio> #include <cstring> int n,m,a[10],arr[10],mark[10]; void dfs(int v){ if(v >= n){ for(int i = 0;i < n;i++) printf("%d ",a[i]); printf("\n"); return ; } for(int i = 0; i< n;i++){ mark[i]=1; dfs(v+1); mark[i] = 0; } } } int main(){ while(scanf("%d",&n)==1){ int i; memset(mark,0,sizeof(mark)); for(i=0;i<n;i++) arr[i]=i+1; dfs(0); } } #include<cstdio> #include<cstring> int n; int a[21],mark[21]; int prime[] = {0,0,1,1,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0,0,0}; //这里是前40的素数表 void dfs(int v) { if(v==n+1) { if(prime[a[v-1]+1])//这里记得判断首尾相加是不是素数 { for(int i=1;i<n;i++) printf("%d ",a[i]); printf("%d",a[n]); printf("\n"); } } else { for(int i=2;i<=n;i++) { if(!mark[i]&&prime[a[v-1]+i]) // 控制相加是素数 { mark[i]=1; a[v]=i; dfs(v+1); mark[i]=0; } } } } int main() { int k=1; while(~scanf("%d",&n)) { printf("Case %d:\n",k++); memset(mark,0,sizeof(mark)); a[1]=1; dfs(2); printf("\n"); } return 0; }
