Anti-prime Sequences
2016-05-27 13:51
302 查看
Anti-prime Sequences
Description
Given a sequence of consecutive integers n,n+1,n+2,...,m, an anti-prime sequence is a rearrangement of these integers so that each adjacent pair of integers sums to a composite (non-prime) number. For example, if n = 1 and m = 10, one such anti-prime sequence is 1,3,5,4,2,6,9,7,8,10. This is also the lexicographically first such sequence.
We can extend the definition by defining a degree danti-prime
sequence as one where all consecutive subsequences of length 2,3,...,d
sum to a composite number. The sequence above is a degree 2 anti-prime
sequence, but not a degree 3, since the subsequence 5, 4, 2 sums to 11.
The lexicographically .rst degree 3 anti-prime sequence for these
numbers is 1,3,5,4,6,2,10,8,7,9.
Input
Input
will consist of multiple input sets. Each set will consist of three
integers, n, m, and d on a single line. The values of n, m and d will
satisfy 1 <= n < m <= 1000, and 2 <= d <= 10. The line 0 0
0 will indicate end of input and should not be processed.
Output
For
each input set, output a single line consisting of a comma-separated
list of integers forming a degree danti-prime sequence (do not insert
any spaces and do not split the output over multiple lines). In the case
where more than one anti-prime sequence exists, print the
lexicographically first one (i.e., output the one with the lowest first
value; in case of a tie, the lowest second value, etc.). In the case
where no anti-prime sequence exists, output
No anti-prime sequence exists.
Sample Input
Sample Output
Time Limit: 3000MS | Memory Limit: 30000K | |
Total Submissions: 3355 | Accepted: 1531 |
Given a sequence of consecutive integers n,n+1,n+2,...,m, an anti-prime sequence is a rearrangement of these integers so that each adjacent pair of integers sums to a composite (non-prime) number. For example, if n = 1 and m = 10, one such anti-prime sequence is 1,3,5,4,2,6,9,7,8,10. This is also the lexicographically first such sequence.
We can extend the definition by defining a degree danti-prime
sequence as one where all consecutive subsequences of length 2,3,...,d
sum to a composite number. The sequence above is a degree 2 anti-prime
sequence, but not a degree 3, since the subsequence 5, 4, 2 sums to 11.
The lexicographically .rst degree 3 anti-prime sequence for these
numbers is 1,3,5,4,6,2,10,8,7,9.
Input
Input
will consist of multiple input sets. Each set will consist of three
integers, n, m, and d on a single line. The values of n, m and d will
satisfy 1 <= n < m <= 1000, and 2 <= d <= 10. The line 0 0
0 will indicate end of input and should not be processed.
Output
For
each input set, output a single line consisting of a comma-separated
list of integers forming a degree danti-prime sequence (do not insert
any spaces and do not split the output over multiple lines). In the case
where more than one anti-prime sequence exists, print the
lexicographically first one (i.e., output the one with the lowest first
value; in case of a tie, the lowest second value, etc.). In the case
where no anti-prime sequence exists, output
No anti-prime sequence exists.
Sample Input
1 10 2 1 10 3 1 10 5 40 60 7 0 0 0
Sample Output
1,3,5,4,2,6,9,7,8,10 1,3,5,4,6,2,10,8,7,9 No anti-prime sequence exists. 40,41,43,42,44,46,45,47,48,50,55,53,52,60,56,49,51,59,58,57,54 题意:在【2,d】长度的连续序列的和都要为合数。 思路:DFS。
#include<stdio.h> #include<algorithm> #include<iostream> #include<stdlib.h> #include<string.h> #include<queue> #include<stack> #include<math.h> using namespace std; typedef long long LL; bool prime[20000]= {0}; int tt[10000]; bool cm[1005]; int ts=0; bool check(int n,int m); int dfs(int n,int m,int d,int kk,int pp); int main(void) { int i,j,k; for(i=2; i<=1000; i++) { if(!prime[i]) { for(j=i; (i*j)<=20000; j++) { prime[i*j]=true; } } } int n,m; while(scanf("%d %d %d",&n,&m,&k),n!=0&&m!=0&&k!=0) { memset(cm,0,sizeof(cm)); ts=0; int uu=dfs(0,m-n+1,k,n,m); if(uu) { printf("%d",tt[0]); for(i=1; i<(m-n+1); i++) { printf(",%d",tt[i]); } printf("\n"); } else printf("No anti-prime sequence exists.\n"); } } bool check(int n,int m) { int i,j; LL sum=tt[m]; for(i=m-1; i>=max(n,0); i--) { sum+=tt[i]; if(!prime[sum]) return false; } return true; } int dfs(int n,int m,int d,int kk,int pp) { int i; if(ts)return 1; if(n==m) { bool cc=check(n-d,m-1); if(!cc) { return 0; } ts=1; return 1; } else { bool cc=check(n-d,n-1); if(cc) { for(i=kk; i<=pp; i++) { if(ts)return 1; if(!cm[i]) { tt =i; cm[i]=true; int uu=dfs(n+1,m,d,kk,pp); cm[i]=false; if(uu)return 1; } } } else return 0; } return 0; }
相关文章推荐
- 你真的了解Instant Run吗?
- D与C#的GUI内存占用比较
- 【EasyUI+MVC-搭建后台框架】
- 瀑布流+UICollectionView
- call by value 和 call by reference
- CEFGlue 加载FLASH 使用HOOK 方式 解决 CEF 加载 PPAPI FLASH 插件时弹出 CMD 命令行 窗口的问题
- easyui datagrid自定义按钮列,即最后面的操作列
- UIScrollView UICollectionView 无法响应touch事件
- UICollectionView 浅析
- UICollectionView cell点击无响应
- 某软件公司的GUID生成代码(GUIDFactory)java+C#
- BBC-The Race and a quiz
- <minigui>GDI双缓冲之道
- leetcode 347. Top K Frequent Elements
- iOS UIImage类扩展(按照位置和大小截图图片中部分图片)
- 序列式容器:vecor,stack,queue用法
- UILabel顶端对齐
- easyui——一些有用的组件使用小记
- NSString(或者说是UILabel)加入 “行间距” 之后的 “高度”计算
- UITextField自定义输入限制(纯数字输入,不输入汉字,小数点后位数限制)