素数环
【摘要】 Prime Ring Problem
Time Limit : 4000/2000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 18 Accepted Submission(s) : 12
F...
Prime Ring Problem
Time Limit : 4000/2000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 18 Accepted Submission(s) : 12
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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.
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.
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)
代码：
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#include <stdio.h>
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#include <string.h>
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int n,ac[21]={1};
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bool in[21];
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bool is_prime(int x)
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{
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for(int i=2;i*i<=x;i++)
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if(x%i==0)
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return false;
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return true;
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}
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void dfs(int cur)
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{
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if(cur == n-1)
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{
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if(is_prime(ac[cur]+1))
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{
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printf("1");
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for(int i=1;i<n;i++)
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printf(" %d",ac[i]);
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printf("\n");
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}
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return ;
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}
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for(int i=2;i<=n;i++)
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{
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if(!in[i] && is_prime(i+ac[cur]))
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{
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in[i]=1;
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ac[cur+1]=i;
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dfs(cur+1);
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in[i]=0;
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}
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}
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}
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int main()
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{
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int ncase=0;
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while(scanf("%d",&n),n)
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{
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printf("Case %d:\n",++ncase);
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memset(in,0,sizeof(in));
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if(n%2==0 || n==1)
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dfs(0);
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else
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printf("No Answer\n");
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}
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return 0;
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}
文章来源: blog.csdn.net，作者：Linux猿，版权归原作者所有，如需转载，请联系作者。
原文链接：blog.csdn.net/nyist_zxp/article/details/9148191
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