Codeforces Round #276 (Div. 1)B. Maximum Value 筛法

D. Maximum Value

 

 

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

Input

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

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

Output

Print the answer to the problem.

Sample test(s)

input

3
3 4 5

output

2

题意：给你一个a数组，问你最大的ai%aj是多少(ai>aj)；
题解：我们要求ai%aj最大，ai%(aj*k)最大，类似筛法去最大就是了

///

#include<bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std ;
typedef long long ll;
typedef unsigned long long ull;
#define mem(a) memset(a,0,sizeof(a))
#define pb push_back
inline ll read()
{
    ll x=,f=;char ch=getchar();
    while(ch<''||ch>''){
        if(ch=='-')f=-;ch=getchar();
    }
    while(ch>=''&&ch<=''){
        x=x*+ch-'';ch=getchar();
    }return x*f;
}
//****************************************
const int  N=+;
#define mod 10000007
#define inf 1000000001
#define maxn 10000

int a[N];
int main() {
    int n=read();
    for(int i=;i<=n;i++) {
        scanf("%d",&a[i]);
      //  H[a[i]]=1;
    }
    sort(a+,a+n+);
    a[]=-;int ans=-;
    for(int i=;i<=n;i++) {
        if(a[i]==a[i-])continue;
        for(int j=a[i]+a[i];j<=a[n];j+=a[i])
        {
            int tmp=lower_bound(a+,a+n+,j)-a;
            ans=max(ans,a[tmp-]%a[i]);
        }
        ans=max(ans,a[n]%a[i]);
    }
    cout<<ans<<endl;
    return ;
}

代码