题目链接:This is the link
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. 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 Sample Input Sample Output Source |
题目大意:
输入m, n, d。求出m,m+1,m+2,````m+n的一个排列。使得任意的连续k个数之和都为合数,2<=k<=d。
输出,按照顺序输出排列,不能组成输出“No anti-prime sequence exists.”
This is the link
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<iostream>
#include <iomanip>
#include<list>
#include<map>
#include<queue>
#include<sstream>
#include<stack>
#include<string>
#include<set>
#include<vector>
using namespace std;
#define PI acos(-1.0)
#define EPS 1e-8
#define MOD 1e9+7
#define LL long long
#define ULL unsigned long long //1844674407370955161
#define INT_INF 0x7f7f7f7f //2139062143
#define LL_INF 0x7f7f7f7f7f7f7f7f //9187201950435737471
const int dr[]={0, 0, -1, 1, -1, -1, 1, 1};
const int dc[]={-1, 1, 0, 0, -1, 1, -1, 1};
// ios.sync_with_stdio(false);
// 那么cin, 就不能跟C的 scanf,sscanf, getchar, fgets之类的一起使用了。
const int N=10005;
int not_prime[N+5];
vector<int > prime;
int num[1005];
int v[1005];
bool falg;
int n,m,d;
void Isprime()
{
not_prime[0]=not_prime[1]=true;
prime.clear();
for(int i=2;i<=N;++i)
{
if(!not_prime[i])
prime.push_back(i);
for(int j=0;j<prime.size()&&i*prime[j]<=N;++j)
{
not_prime[i*prime[j]]=true;
if(!(i%prime[j]))
break;
}
}
}
bool judge(int k,int d)
{
for(int i=d;i>=2;--i)
{
if(k-i>=0)
{
int ans=0;
for(int j=k-i+1;j<=k;++j)
ans+=num[j];
if(!not_prime[ans])//表示为素数
return false;
}
}
return true;
}
void dfs(int k)
{
if(falg)
return ;
if(k==m-n+2)
{
printf("%d",num[1]);
for(int i=2;i<=m-n+1;++i)
printf(",%d",num[i]);
printf("\n");
falg=true;
return ;
}
for(int i=n;i<=m;++i)
{
if(v[i])
{
num[k]=i;
if(judge(k,d))
{
v[i]=false;
dfs(k+1);
v[i]=true;
}
}
}
}
int main()
{
Isprime();
while(scanf("%d%d%d",&n,&m,&d)!=EOF)
{
if(!n&&!m&&!d)
break;
memset(v,true,sizeof(v));
falg=false;
dfs(1);
if(!falg)
printf("No anti-prime sequence exists.\n");
}
return 0;
}