CF-1176D

Posted on | In | Views:

题意

Screen Shot 2019-06-12 at 5.36.51 PM

题解

排序, 从大到小处理, 若当前最大的为素数, 那它一定是 $b$, 若当前最大数为合数, 那它一定是 $a$

代码

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
#include <bits/stdc++.h>
using namespace std;
#define forl(i, l, r) for (int i = l; i <= r; i++)
#define forr(i, r, l) for (int i = r; i >= l; i--)
#define for1(i, n) for (int i = 1; i <= n; i++)
#define fro1(i, n) for (int i = 1; i <= n; i++)
#define for0(i, n) for (int i = 0; i < n; i++)
#define fro0(i, n) for (int i = 0; i < n; i++)
#define meminf(a) memset(a, inf, sizeof(a))
#define mem_1(a) memset(a, -1, sizeof(a))
#define mem0(a) memset(a, 0, sizeof(a))
#define memcp(a,b) memcpy(a,b,sizeof(b))
#define oper(type) bool operator <(const type y)const
#define mp make_pair
#define pu_b push_back
#define pu_f push_front
#define po_b pop_back
#define po_f pop_front
#define fi first
#define se second
#define whiel while
#define retrun return
typedef pair<long long, long long> pll;
typedef vector<long long> vll;
typedef pair<int, int> pii;
typedef unsigned long long ull;
typedef vector<int> vii;
typedef long double db;
typedef long long ll;
typedef int itn;
int in(int &a,int &b,int &c,int &d){return scanf("%d%d%d%d",&a,&b,&c,&d);}
int in(int &a,int &b,int &c){return scanf("%d%d%d",&a,&b,&c);}
int in(int &a,int &b){return scanf("%d%d",&a,&b);}
int in(ll &a){return scanf("%lld",&a);}
int in(int &a){return scanf("%d",&a);}
int in(char *s){return scanf("%s",s);}
int in(char &s){return scanf("%c",&s);}
int in(db &a){return scanf("%Lf",&a);}
void out(int a){printf("%d ",a);}
void outln(int a){printf("%d\n",a);}
void out(ll a){printf("%lld ",a);}
void outln(ll a){printf("%lld\n",a);}
const db pi = acos((db)-1);
const ll inf =0x3f3f3f3f;
const db eps = 1e-8;
const int N = 3e6;
const ll mod = 1e9+7;
int sign(db a) { return a < -eps ? -1 : a > eps;}
int db_cmp(db a, db b){ return sign(a-b); }

int Div[N];
int prime[N],num_prime=0;
void get_prime(){
  	
    for(int i=2;i<N;i++){
        if(!Div[i]) {
            prime[num_prime++]=i;
            Div[i]=-num_prime;
        }
        for(int j=0;j<num_prime&&(ll)i*prime[j]<N;j++){
            int k = i*prime[j];
            Div[k] = i;
            if(i % prime[j] == 0) break;
        }
    }
}
itn n,a[400005],cnt[N];
int main() {
    get_prime();
    in(n);
    for0(i,2*n){
        in(a[i]);
        cnt[a[i]]++;
    }
    sort(a,a+2*n);
    for(int i=2*n-1;i>=0;i--){
        if(cnt[a[i]]){
            if(Div[a[i]]>0){
                out(a[i]);
                cnt[a[i]]--;
                cnt[Div[a[i]]]--;
            }else{
                int tmp=-Div[a[i]];
                out(tmp);
                cnt[tmp]--;
                cnt[a[i]]--;
            }
        }
    }
    return 0;
}