CodeFores 665D Simple Subset(贪心)
CodeFores 665D Simple Subset(贪心)
D. Simple Subset
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
A tuple of positive integers {x1, x2, ..., xk} is called simple if for all pairs of positive integers (i, j) (1 ≤ i < j ≤ k), xi + xj is a prime.
You are given an array a with n positive integers a1, a2, ..., an (not necessary distinct). You want to find a simple subset of the array awith the maximum size.
A prime number (or a prime) is a natural number greater than 1 that has no positive divisors other than 1 and itself.
Let's define a subset of the array a as a tuple that can be obtained from a by removing some (possibly all) elements of it.
Input
The first line contains integer n (1 ≤ n ≤ 1000) — the number of integers in the array a.
The second line contains n integers ai (1 ≤ ai ≤ 106) — the elements of the array a.
Output
On the first line print integer m — the maximum possible size of simple subset of a.
On the second line print m integers bl — the elements of the simple subset of the array a with the maximum size.
If there is more than one solution you can print any of them. You can print the elements of the subset in any order.
Examples
input
2
2 3output
2
3 2input
2
2 2output
1
2input
3
2 1 1output
3
1 1 2input
2
83 14output
2
14 83
首先对于每个数，找出来和它的和不是素数的数，并统计个数。然后贪心的把个数最大的那个数删除，同时，和它相关的那些数的个数就少了1，然后再选取个数最大的，直到所有数的个数为0.用了优先队列来实现。一开始的时候要把重复的数字合并起来，要不然这个方法会超时。#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <math.h>
#include <stdio.h>
#include <vector>
#include <queue>
using namespace std;
#define MAX 1000
typedef long long int LL;
vector<int> a[MAX+5];
int b[MAX+5];
int c[MAX+5];
int tag[MAX+5];
bool flag[MAX+5];
bool t[MAX*1000+5];
int n;
struct Node
{
int pos;
int value;
Node(){};
Node(int pos,int value){this->pos=pos;this->value=value;}
friend bool operator<(Node a,Node b){return a.value<b.value;}
};
priority_queue<Node>q;
bool isPrime(LL x)
{
if(x==1) return 0;
if(x==2) return 1;
for(int i=2;i*i<=x;i++)
{
if(x%i==0)
return 0;
}
return 1;
}
int main()
{
while(scanf("%d",&n)!=EOF)
{
for(int i=1;i<=n;i++)
scanf("%d",&c[i]);
memset(t,false,sizeof(t));
int cnt=0;
for(int i=1;i<=n;i++)
{
if(!t[c[i]]||c[i]==1)
{
b[++cnt]=c[i];
t[c[i]]=true;
}
}
n=cnt;
memset(tag,0,sizeof(tag));
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
{
if(i==j) continue;
if(!isPrime(b[i]+b[j]))
{a[i].push_back(j);tag[i]++;}
}
}
for(int i=1;i<=n;i++)
q.push(Node(i,tag[i]));
memset(flag,true,sizeof(flag));
int num=n;
while(!q.empty())
{
Node term=q.top();
q.pop();
if(term.value!=tag[term.pos]) continue;
if(term.value==0)
break;
flag[term.pos]=false;num--;
for(int i=0;i<a[term.pos].size();i++)
{
if(flag[a[term.pos][i]]==false) continue;
tag[a[term.pos][i]]--;
q.push(Node(a[term.pos][i], tag[a[term.pos][i]]));
}
}
printf("%d\n",num);
for(int i=1;i<=n;i++)
{
if(flag[i])
{
if(i==n)
printf("%d\n",b[i]);
else
printf("%d ",b[i]);
}
}
}
return 0;
}