Codeforces 485D Maximum Value【思维+数论】

D. Maximum Value

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a sequence a consisting of n integers. Find the maximum possible value of  (integer remainder of a_i divided bya_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.

Examples

Input

33 4 5

Output

2

题目大意：

给你N个数，让你找到两个数Ai，Aj，使得Ai%Aj的值最大，这里需要保证Ai>Aj；

题解:排序后,枚举每个数a[i],然后再枚举a[i]的倍数k*a[i],找到k*a[i]~(k+1)*a[i]中模上a[i]的最大的数,那个数一定是小于(k+1)*a[i]的最大数,然后取模后与答案比较,取最大值

#include<bits/stdc++.h>#define inf 0x3f3f3f3fusing namespace std;typedef long long ll;typedef pair<int,int> pii;const int N=2000010,MOD=1e9+7;int a[200020];int pre[N];bool have[N];int main(){    int n;cin>>n;    for(int i=0;i<n;i++) scanf("%d",a+i),have[a[i]]=1;    sort(a,a+n);    n = unique(a,a+n)-a;    int mx = a[n-1];    mx<<=1;    int last=0;    for(int i=1;i<=mx;i++){        if(!have[i]) pre[i] = last;        else{            pre[i]=last;            last=i;        }    }    int ans=0;    int i=a[0]==1;    for(;i<n;i++){        for(int j=a[i]+a[i];j<mx;j+=a[i]){            if(j-a[n-1]>=a[i]) break;            int ret = pre[j];            if(j-ret >= a[i]) continue;            ans = max(ans,a[i]-(j-ret));        }    }    cout<<ans<<endl;    return 0;}