Codeforces Round #276 (Div. 一) B. Maximum Value(哈兮）
Codeforces Round #276 (Div. 1) B. Maximum Value(哈兮）
You are given a sequence a consisting ofn integers. Find the maximum possible value of
(integer remainder ofai divided byaj), where1 ≤ 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 integersai (1 ≤ ai ≤ 106).
Output
Print the answer to the problem.
Sample test(s)
Input
3 3 4 5
Output
2
题目大意：求解最大的a[j]%a[i]
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#define inf 2000000+10
int ha[inf];
using namespace std;
int main()
{
int n,m,ma=0,i,j;//若求最大的a[j]%a[i]的数只有可能在a[j]倍数小的数中产生
scanf("%d",&n);
memset(ha,-1,sizeof(ha));
for(i=0; i<n; i++)//哈兮预处理一下
{
scanf("%d",&m);
ha[m]=m;
}
for(i=1; i<inf; i++)//2*10^6在1000ms不会超
if(ha[i]!=i)//当前的哈兮数组为-1,则赋值为最大乡邻的最大数
{
ha[i]=ha[i-1];
}
int ans=0;
for(i=2;i<inf;i++)
{
if(ha[i]==i)
{
for(j=i*2-1;j<inf;j+=i)//最大的数在倍数的减-1中产生
{
ans=max(ans,ha[j]%i);
}
}
}
printf("%d\n",ans);
return 0;
}
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