Codeforces Round #206 (Div. 2) E-Vasya and Beautiful Arrays

来源：互联网发布：萨克斯伴奏软件编辑：程序博客网时间：2024/04/29 11:12

Vasya's got a birthday coming up and his mom decided to give him an array of positive integers a of length n.

Vasya thinks that an array's beauty is the greatest common divisor of all its elements. His mom, of course, wants to give him as beautiful an array as possible (with largest possible beauty). Unfortunately, the shop has only one array a left. On the plus side, the seller said that he could decrease some numbers in the array (no more than by k for each number).

The seller can obtain array b from array a if the following conditions hold: bi > 0; 0 ≤ ai - bi ≤ k for all 1 ≤ i ≤ n.

Help mom find the maximum possible beauty of the array she will give to Vasya (that seller can obtain).

Input

The first line contains two integers n and k (1 ≤ n ≤ 3·105; 1 ≤ k ≤ 106). The second line contains n integers ai (1 ≤ ai ≤ 106) — arraya.

Output

In the single line print a single number — the maximum possible beauty of the resulting array.

Examples
input
6 13 6 10 12 13 16
output
3
input
5 38 21 52 15 77
output
7

Note

In the first sample we can obtain the array:

3 6 9 12 12 15

In the second sample we can obtain the next array:

7 21 49 14 77

题意：首先输入两个数N和K然后输入N个数，让我们找到最大的数例如m使得ai mod m小于等于K；

思路：首先我们知道m必定小于等于N个数中最小的，直接由最小a0开始逐个遍历找m肯定超时，那么举个例子 9 11 24 37 52， K=2，

当m=9时24不行，m=8，11不行，m=7, 11还不行，m=6, 11还不行，m=5， 11可以了，由8到5，11都不行，若是像之前那般遍历

就是减了多余两次，所以想到怎么让8直接变成5来缩短次数提高效率，11/(11/8+1)。

同附上Codeforces题解代码。题解可去Codeforces查看。

下面附上代码：

#include <bits/stdc++.h>using namespace std;#define INF 0x3f3f3f3fint a[300005];int main(){  int n,k;  scanf("%d%d",&n,&k);  for(int i=0;i<n;i++)  {    scanf("%d",&a[i]);  }  sort(a,a+n);  int ans=a[0];  while(1)  {    int flag=0;    for(int i=0;i<n;i++)    {      if(a[i]%ans>k)      {        ans=a[i]/(a[i]/ans+1);        flag=1;        break;      }    }    if(flag==0)    {      printf("%d\n",ans);      break;    }  }}

Codeforces题解代码：

#include <bits/stdc++.h>using namespace std;#define INF 0x3f3f3f3fint a[300005];int main(){  int n,k;  scanf("%d%d",&n,&k);  for(int i=0;i<n;i++)  {    scanf("%d",&a[i]);  }  sort(a,a+n);  if(a[0]<=k+1)  {    printf("%d\n",a[0]);  }  else  {    int x1,x2,sum=0,i;    for(i=a[0];i>k;i--)    {      int ans=a[n-1];      sum=0;      for(int j=1;i*j<=ans;j++)       {        x1=lower_bound(a,a+n,j*i)-a;        x2=lower_bound(a,a+n,j*i+k+1)-a; //注意是j*i+k+1不是j*i+k        sum+=(x2-x1);      }      if(sum==n)      {        break;      }    }    printf("%d\n",i);  }}

0 0