CodeForces - 484BMaximum Value（hash优化）

Posted 2020-10-15 余生漫漫浪

个人心得：周测题目，一题没出，难受得一批。这个题目做了一个半小时还是无限WR，虽然考虑到了二分答案这个点上面了，

奈何二分比较差就想用自己的优化，虽然卡在了a=k*b+c，这里但是后面结束了这样解决还是超时了，看了一下网上的hash，思想一样

但是却优化了很多，服气

题目：

You are given a sequence a consisting of n integers. Find the maximum possible value of  (integer remainder of a_i divided by a_j), where 1?≤?i,?j?≤?n and a_i?≥?a_j.

Input

The first line contains integer n — the length of the sequence (1?≤?n?≤?2·10⁵).

The second line contains n space-separated integers a_i (1?≤?a_i?≤?10⁶).

Output

Print the answer to the problem.

Example

Input

3
3 4 5

Output

2

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cmath>
 4 #include<cstring>
 5 #include<iomanip>
 6 #include<algorithm>
 7 using namespace std;
 8 #define inf 1<<29
 9 #define infa 2000000+10
10 #define nu 50005
11 int n,m;
12 int book[infa];
13 int main()
14 {
15     scanf("%d",&n);
16     memset(book,-1,sizeof(book));
17     int maxnum;
18     for(int i=0;i<n;i++){
19        scanf("%d",&m);
20         book[m]=m;
21     }
22     for(int i=1;i<infa; i++)
23         if(book[i]!=i)
24         {
25             book[i]=book[i-1];
26         }
27         int sum=0;
28         for(int i=1;i<infa;i++)
29         {
30             if(book[i]==i)
31                 {
32                     for(int j=i*2-1;j<infa;j+=i)
33                         sum=max(sum,book[j]%i);
34                 }
35         }
36            cout<<sum<<endl;
37     return 0;
38 }

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