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 

 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 }

View Code