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;
}
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