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[2*M+10];
int main()
{
    int n,i,j,k,ans;
    cin>>n>>k;
    for(i = 1; i <= n ; i++)
    {
        cin>>a[i];
        vis[a[i]]++;
    }
    sort(a+1,a+n+1);
    for(i = 1 ;i <= M*2 ;i++)
    vis[i] += vis[i-1];
    if(k>=a[1])
    {
        printf("%d
",a[1]);
        return 0;
    }
    for(i = k ; i <= a[n] ; i++)
    {
        int sum=0;
        for(j = i ; j <= a[n] ; j+=i)
        {
            sum+=vis[j+k]-vis[j-1];
        }
        if(sum==n)
        ans = i;
    }
    cout<<ans<<endl;
    return 0;
}