HDU Prime Ring Problem (DFS+素数打表)
2015-08-14 14:40
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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
Sample Output
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
#include<iostream> #include<cstdio> #include<cstring> using namespace std; int n,vis[40],ans[40],prime[100]; void init_prime() { int i,j; for(i=2;i<100;i++) { if(!prime[i]){ for(j=2*i;j<=100;j+=i) prime[j]=1; } } } void dfs(int num,int cnt) { int i; if(cnt==n) { for(i=0;i<n;i++) { if(i==0) printf("%d",ans[i]); else printf(" %d",ans[i]); } printf("\n"); } for(i=2;i<=n;i++) { if(vis[i]==1) continue; if(prime[i+num]==0) { if(cnt==n-1 && prime[i+1]) continue; vis[i]=1; ans[cnt]=i; dfs(i,cnt+1); vis[i]=0; } } } int main() { int number,i,j; number=0; init_prime(); while(scanf("%d",&n)!=EOF){ printf("Case %d:\n",++number); memset(vis,0,sizeof(vis)); ans[0]=1; vis[1]=1; dfs(1,1); printf("\n"); } return 0; }
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