Maximum Value(哈希)

B. 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 ai divided by aj), 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

Hash记录输入的数值,Dp[i]记录输入中距离i最近的数值(不包括本身)

这里写代码片#include <set>
#include <map>
#include <list>
#include <stack>
#include <cmath>
#include <queue>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#define PI cos(-1.0)
#define RR freopen("input.txt","r",stdin)

using namespace std;

typedef long long LL;

const int MAX = 2*1e6+10;

const int R =1e6;

int Dp[MAX];

int n;

int a[MAX];

bool Hash[MAX];

int main()
{
    int n;
    memset(Hash,false,sizeof(Hash));
    scanf("%d",&n);
    int Min=MAX,Max=0;
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
        Hash[a[i]]=true;
        Min=min(Min,a[i]);
        Max=max(Max,a[i]);
    }
    for(int i=Min;i<MAX;i++)
    {
        if(Hash[i-1])
        {
            Dp[i]=i-1;
        }
        else
        {
            Dp[i]=Dp[i-1];
        }
    }
    int ans=0;
    for(int i=Min;i<=R;i++)
    {
        if(Hash[i])
        {
            for(int j=i*2;;j+=i)
            {
                if(Dp[j]<i)
                {
                    continue;
                }
                ans=max(Dp[j]%i,ans);
                if(Dp[j]==Max)
                {
                    break;
                }
            }
        }
    }
    printf("%d\n",ans);
    return 0;
}