Codeforces-959D Mahmoud and Ehab and another array construction task(数论,质因子分解)

Mahmoud has an array a consisting of n integers. He asked Ehab to find another array b of the same length such that:

b is lexicographically greater than or equal to a.
bi ≥ 2.
b is pairwise coprime: for every 1 ≤ i < j ≤ n, bi and bj are coprime, i. e. GCD(bi, bj) = 1, where GCD(w, z) is the greatest common divisor of w and z.

Ehab wants to choose a special array so he wants the lexicographically minimal array between all the variants. Can you find it?

An array x is lexicographically greater than an array y if there exists an index i such than xi > yi and xj = yj for all 1 ≤ j < i. An array x is equal to an array y if xi = yi for all 1 ≤ i ≤ n.

Input

The first line contains an integer n (1 ≤ n ≤ 105), the number of elements in a and b.

The second line contains n integers a1, a2, ..., an (2 ≤ ai ≤ 105), the elements of a.

Output

Output n space-separated integers, the i-th of them representing bi.

Note

Note that in the second sample, the array is already pairwise coprime so we printed it.

题意：

给定数组a，求大于或者等于数组a字典序的数组b，满足任意i,j，gcd(bi, bj) = 1。并且是在所有满足条件中的字典序最小的一个。

题解:

对于第i个数先检验a[i]是否符合标准,若不符合,则后面的数直接找未选取的符合标准的最小素数,判断是否互质选择的做法是将之前的数质因子分解,标记质因子,此处比较麻烦的地方在于需要判断该数是否选取,若选取则将质因子标记,若不选取则不标记质因子。

#include<stdio.h>
#include<algorithm>
#include<iostream>
#include<string.h>
#include<vector>
#include<stdlib.h>
#include<math.h>
#include<queue>
#include<deque>
#include<ctype.h>
#include<map>
#include<set>
#include<stack>
#include<string>
#include<algorithm>
#define gcd(a,b) __gcd(a,b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define FAST_IO ios::sync_with_stdio(false)
#define mem(a,b) memset(a,b,sizeof(a))
const double PI = acos(-1.0);
const double eps = 1e-6;
const int MAX=1e5+10;
const long long INF=0x7FFFFFFFFFFFFFFFLL;
const int inf=0x3f3f3f3f;
const int mod=1e9+7;
typedef long long ll;
using namespace std;
const int N=3e6;
int a[N+5],ans[N+5],vis[N+5],prime[N+5],cot,isprime[N+5];
void primeall()
{
    for(int i=2;i<=sqrt(N);i++)
        if(!isprime[i])
            for(int j=i*i;j<=N;j+=i)
                isprime[j]=1;

    for(int i = 2; i <= N; ++i)
        if(!isprime[i])
            prime[++cot] = i;


}

int judge(int id,int flag)
{
    for(int i=1;i<=cot && prime[i]*prime[i]<=id;i++)
    {
        if(!flag && id%prime[i]==0 &&  vis[prime[i]])
                return 0;
        if( flag && id%prime[i]==0) vis[prime[i]]=1;

        while(id%prime[i]==0)
            id/=prime[i];
    }
    if(id>1)
    {
        if(vis[id]) return 0;
        if(flag) vis[id]=1;
    }
    return 1;
}

int work(int id)
{
    if(judge(a[id],0))
    {
        judge(a[id],1);
        ans[id]=a[id];
        return 0;
    }

    for(int i=a[id]+1;;i++)
        if(judge(i,0))
        {
            judge(i,1);
            ans[id]=i;
            break;
        }
    return 1;
}
int main()
{
    primeall();
    int n;
    cin>>n;

    int flag=0,no=1;
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
        if(flag)
        {
            while(vis[prime[no]])
                no++;
            ans[i]=prime[no++];
            continue;
        }
        if(work(i))
            flag=1;
    }

    for(int i=1;i<=n;i++)
        printf("%d ",ans[i]);
    return 0;
}

Codeforces-959D Mahmoud and Ehab and another array construction task(数论,质因子分解)

猜你喜欢