Prefix Sum Primes CodeForces - 1150C(思维)

We’re giving away nice huge bags containing number tiles! A bag we want to present to you contains nn tiles. Each of them has a single number written on it — either 11 or 22.

However, there is one condition you must fulfill in order to receive the prize. You will need to put all the tiles from the bag in a sequence, in any order you wish. We will then compute the sums of all prefixes in the sequence, and then count how many of these sums are prime numbers. If you want to keep the prize, you will need to maximize the number of primes you get.

Can you win the prize? Hurry up, the bags are waiting!

Input
The first line of the input contains a single integer nn (1≤n≤2000001≤n≤200000) — the number of number tiles in the bag. The following line contains nn space-separated integers a1,a2,…,ana1,a2,…,an (ai∈{1,2}ai∈{1,2}) — the values written on the tiles.

Output
Output a permutation b1,b2,…,bnb1,b2,…,bn of the input sequence (a1,a2,…,an)(a1,a2,…,an) maximizing the number of the prefix sums being prime numbers. If there are multiple optimal permutations, output any.

Examples
Input
5
1 2 1 2 1
Output
1 1 1 2 2
Input
9
1 1 2 1 1 1 2 1 1
Output
1 1 1 2 1 1 1 2 1
Note
The first solution produces the prefix sums 1,2,3,5,71,2,3,5,7 (four primes constructed), while the prefix sums in the second solution are 1,2,3,5,6,7,8,10,111,2,3,5,6,7,8,10,11 (five primes). Primes are marked bold and blue. In each of these cases, the number of produced primes is maximum possible.
题意:用1和2构造出前缀和素数最多的数列。

我们要知道,除了2和3之间的差为1外,其他两个素数之间的差值一定是一个偶数,即2的倍数,所以贪心,拼出3后,优先填2,再填1.

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define Fin             freopen("in.txt","r",stdin)
#define Fout            freopen("out.txt","w",stdout)
#define Case(T)         int T;for(scanf("%d",&T);T--;)
#define fo(i,a,b)              for(int i = a; i < b; ++i)
#define fd(i,a,b)              for(int i = a; i >= b; --i)
#define me(a,b) memset(a,b,sizeof(a))
#define fi(a,n,val)    fill(a,a+n,val)
#define Scand(n)       scanf("%d",&n)
#define Scand2(a,b)     scanf("%d%d",&a,&b)
#define Scand3(a,b,c)     scanf("%d%d%d",&a,&b,&c)
#define Scand4(a,b,c,d)     scanf("%d%d%d%d",&a,&b,&c,&d)
#define Scans(s)       scanf("%s",s)
#define random(a,b)    a+rand()%(b-a+1)
using namespace std;
typedef long long ll;
ll gcd(ll a,ll b) { return b ? gcd(b,a%b): a; }
const int maxn = 2e5 + 50;
const int INFint = 0xffffff;
const ll INFll = (ll)1e18;
 
#ifndef ONLINE_JUDGE
 
#endif // ONLINE_JUDGE
 
inline int readint(){
    int sgn = 1; int sum = 0;
    char ch = getchar();
    while(!('0' <= ch && ch <= '9')){
        if(ch == '-')   sgn = -sgn;
        ch = getchar();
    }
    while('0' <= ch && ch <= '9'){
        sum = sum * 10 + (ch - '0');
        ch = getchar();
    }
    return sgn * sum;
}
 
inline ll readll(){
    ll sgn = 1; ll sum = 0;
    char ch = getchar();
    while (ch < '0' || ch > '9') {
        if(ch == '-')   sgn = -sgn;
        ch = getchar();
    }
    while ('0' <= ch && ch <= '9') {
        sum = sum*10+(ch-'0');
        ch = getchar();
    }
    return sgn*sum;
}
 
int n;
int arr[maxn];
int cnt1, cnt2;
 
int main()
{
    #ifndef ONLINE_JUDGE
        //Fin;
    #endif // ONLINE_JUDGE
    cnt1 = cnt2 = 0;
    n = readint();
    for(int i = 0; i < n; ++i){
        arr[i] = readint();
        if(arr[i]%2)
            cnt1++;
        else
            cnt2++;
    }
    if(cnt2){
        cout << 2;
        cnt2--;
    }
    if(cnt1){
        cout << ' ' << 1;
        cnt1--;
    }
    while(cnt2){
        cout << ' ' << 2;
        cnt2--;
    }
    while(cnt1){
        cout << ' ' << 1;
        cnt1--;
    }
    return 0;
}

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