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

时间:2023-08-25 16:43:02

B. Maximum Value

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/484/problem/B

Description

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 Input

3
3 4 5

Sample Output

2

HINT

题意

给你n个数(n<=1e5),每个数(大小<=2*1e6),要求找到最大的ai%aj(ai>aj)

题解:

类似于筛法,对于每个数,我们都筛出他的倍数

ai%aj最大,可以转化为ai*k-aj最小,记录下每个数的倍数的前面的最大的数就好了

代码

#include<iostream>
#include<stdio.h>
#include<math.h>
using namespace std; int pre[]; int main()
{
int n;scanf("%d",&n);
for(int i=;i<=n;i++)
{
int x;scanf("%d",&x);
pre[x]=x;
}
for(int i=;i<=;i++)
pre[i]=max(pre[i],pre[i-]);
int ans = ;
int flag = ;
for(int i=;i<=;i++)
{
if(pre[i]==i)
for(int j=i;j<=;j+=i)
ans = max(pre[i+j-]%i,ans);
}
printf("%d\n",ans);
}