Codeforces Round #206 div1 C

CF的专业题解 ：

The problem was to find greatest d, such that a_i ≥ d,  a_i mod d ≤ k holds for each i.

Let m = min(a_i), then d ≤ m. Let consider two cases:

. In this case we will brute force answer from k + 1 to m. We can check, if number d is a correct answer in the following way:

We have to check that a_i mod d ≤ k for some fixed d, which is equals to , where . Since all these intervals [x·d..x·d + k] doesn't intersects each with other, we can just check that , where cnt[l..r] — count of numbers a_i in the interval [l..r].

我用汉语大体概括一下

对于 k值是大于等于数组里面的最小值a1的时候  那么肯定答案就是a1

else 就暴力枚举可能的答案d  对于可以减去[0-k] 把d整除的区间有 [d,..d+k] [2d..2d+k] [3d..3d+k]等等。。然后可以标记数组里出现的数 数下在这些区间出现的数有多少个 如果等于N 那么就是满足条件的

 #include <iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<stdlib.h>
 #include<vector>
 #include<queue>
 #include<cmath>
 using namespace std;
 #define N 300010
 #define M 1000000
 int a[N],vis[*M+];
 int main()
 {
     int n,i,j,k,ans;
     cin>>n>>k;
     for(i = ; i <= n ; i++)
     {
         cin>>a[i];
         vis[a[i]]++;
     }
     sort(a+,a+n+);
     for(i =  ;i <= M* ;i++)
     vis[i] += vis[i-];
     if(k>=a[])
     {
         printf("%d\n",a[]);
         return ;
     }
     for(i = k ; i <= a[n] ; i++)
     {
         int sum=;
         for(j = i ; j <= a[n] ; j+=i)
         {
             sum+=vis[j+k]-vis[j-];
         }
         if(sum==n)
         ans = i;
     }
     cout<<ans<<endl;
     return ;
 }