Codeforces Round #206 div1 C-优快云博客

本文讨论了一个关于在数组中找到满足特定条件的最大整数的问题。首先，解释了问题的背景和基本概念，接着介绍了如何通过枚举和区间计数的方法解决这个问题。详细展示了算法的实现过程，包括初始化、排序、区间计数以及最终找到满足条件的最大整数。

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:

$\text{[math]}$

$\text{[math]}$ . 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 $\text{[math]}$ , where $\text{[math]}$ . Since all these intervals [x·d..x·d + k] doesn't intersects each with other, we can just check that $\text{[math]}$ , 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

 1 #include <iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #include<stdlib.h>
 6 #include<vector>
 7 #include<queue>
 8 #include<cmath>
 9 using namespace std;
10 #define N 300010
11 #define M 1000000
12 int a[N],vis[2*M+10];
13 int main()
14 {
15     int n,i,j,k,ans;
16     cin>>n>>k;
17     for(i = 1; i <= n ; i++)
18     {
19         cin>>a[i];
20         vis[a[i]]++;
21     }
22     sort(a+1,a+n+1);
23     for(i = 1 ;i <= M*2 ;i++)
24     vis[i] += vis[i-1];
25     if(k>=a[1])
26     {
27         printf("%d\n",a[1]);
28         return 0;
29     }
30     for(i = k ; i <= a[n] ; i++)
31     {
32         int sum=0;
33         for(j = i ; j <= a[n] ; j+=i)
34         {
35             sum+=vis[j+k]-vis[j-1];
36         }
37         if(sum==n)
38         ans = i;
39     }
40     cout<<ans<<endl;
41     return 0;
42 }