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

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.

[b]Input[/b]

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).

[b]Output[/b]

Print the answer to the problem.

[b]Sample Input[/b]

3
3 4 5

[b]Sample Output[/b]

2

HINT

[b]题意
[/b]

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)

[b]题解：[/b]

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

[b]代码[/b]

#include<iostream>
#include<stdio.h>
#include<math.h>
using namespace std;

int pre[4060005];

int main()
{
int n;scanf("%d",&n);
for(int i=1;i<=n;i++)
{
int x;scanf("%d",&x);
pre[x]=x;
}
for(int i=1;i<=4000005;i++)
pre[i]=max(pre[i],pre[i-1]);
int ans = 0;
int flag = 1;
for(int i=1;i<=2000005;i++)
{
if(pre[i]==i)
for(int j=i;j<=2000005;j+=i)
ans = max(pre[i+j-1]%i,ans);
}
printf("%d\n",ans);
}