A - Minimum Modular

最新推荐文章于 2024-10-22 11:13:32 发布

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最新推荐文章于 2024-10-22 11:13:32 发布

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本文链接：https://blog.youkuaiyun.com/qq_41694395/article/details/79417385

本文探讨了一道算法题目，给定n个不同的整数，在移除最多k个整数后，寻找使得任意两个剩余整数之差的模数不小于该模数的最小正整数m。通过输入输出示例及提供的C++代码实现了解决方案。

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You have been given n distinct integers a₁, a₂, ..., a_n. You can remove at most k of them. Find the minimum modular m (m > 0), so that for every pair of the remaining integers (a_i, a_j), the following unequality holds: $\text{[math]}$ .

Input

The first line contains two integers n and k (1 ≤ n ≤ 5000, 0 ≤ k ≤ 4), which we have mentioned above.

The second line contains n distinct integers a₁, a₂, ..., a_n (0 ≤ a_i ≤ 10⁶).

Output

Print a single positive integer — the minimum m.

Example

Input

7 0
0 2 3 6 7 12 18

Output

Input

7 1
0 2 3 6 7 12 18

Output

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <cctype>
#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <queue>
#include <map>
#include <set>
#include <sstream>
#include<iomanip>
using namespace std;
typedef long long ll;
#define inf int(0x3f3f3f3f)
#define mod int(1e9+7)
#define pi acos(-1)

const int maxn = 1e6 + 5;

int a[maxn],n,k,re[maxn],h[maxn];

int main()
{
    int ma=-1;
    scanf("%d%d",&n,&k);
    memset(re,0,sizeof(re));
    memset(h,0,sizeof(h));
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
        if(ma<a[i]) ma=a[i];
    }
  for(int i=1;i<=n;i++)
  {
      for(int j=1;j<i;j++)
      {
          int tmp=a[i]-a[j]>0?a[i]-a[j]:a[j]-a[i];
          h[tmp]++;
      }
  }
  int res,flag=1;
  for(int m=1;m<=maxn+1;m++)
  {
      int tk=k;
      int sum=0;
      flag=1;
      for(int i=m;i<=1e6;i+=m)
      {
          sum+=h[i];
          if(sum>k*(k+1)/2) break;
      }
      if(sum>k*(k+1)/2) continue;
      res=m;
      for(int i=1;i<=n;i++)
      {
          if(!re[a[i]%res]) re[a[i]%res]++;
          else
            {
                tk--;
                if(tk<0) {flag=0;break;}
            }
      }
      memset(re,0,sizeof(re));
      if(flag) break;
  }
  printf("%d\n",res);
}