标签：else ISE size fir memset names std scan strong
A ring is composed of n (even number) circles as shown in diagram. Put natural numbers 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 ≤ 16) 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. You are to write a program that completes above process. 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
1 #include <cstdio> 2 #include <iostream> 3 #include <queue> 4 #include <vector> 5 #include<string.h> 6 #include<map> 7 #include<bits/stdc++.h> 8 #define LL long long 9 #define maxn 1005 10 using namespace std; 11 int n,ans,tot; 12 int a[20],check[100]; 13 bool vis[20]; 14 int isprime(int x) 15 { 16 for(int i=2;i*i<=x;i++) 17 if(x%i==0)return 0; 18 return 1; 19 } 20 void dfs(int cur) 21 { 22 if(cur==n+1&&check[a[1]+a[n]]) 23 { 24 for(int i=1;i<=n;i++) 25 { 26 if(i!=1) 27 printf(" %d",a[i]); 28 else 29 printf("%d",a[i]); 30 } 31 printf("\n"); 32 return; 33 } 34 for(int i=2;i<=n;i++) 35 { 36 if(!vis[i]&&check[i+a[cur-1]]) 37 { 38 a[cur]=i; 39 vis[i]=1; 40 dfs(cur+1); 41 vis[i]=0; 42 } 43 } 44 } 45 46 int main() 47 { 48 ans=1; 49 while(~scanf("%d",&n)) 50 { 51 if(ans>1)printf("\n"); 52 memset(vis,0,sizeof vis); 53 for(int i=2;i<=2*n;i++) 54 check[i]=isprime(i); 55 a[1]=1;//题目要求从1开始逆时针排列 56 vis[1]=1; 57 printf("Case %d:\n",ans++); 58 dfs(2); 59 //printf("\n");这样写的话会PE，原因是在最后一次输出时会多空一行 60 } 61 return 0; 62 }
标签：else ISE size fir memset names std scan strong
原文地址：https://www.cnblogs.com/zuiaimiusi/p/10888029.html
