CodeForces 484B Maximum Value (数学，其实我也不知道咋分类)

You are given a sequence a consisting of n integers. Find the maximum possible value of $\text{[math]}$ (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

题意：给一堆数，然后要求找两个数，大的数除小的数求余，求最大的余值是多少

思路：类似筛表法的思想，对于每个aj找ai。对每个aj，就for(p=aj;p<mx;p+=aj) 然后二分一下找小于a[p]且最接近a[p]的数，然后用这个数维护答案。另外一点由于这里的数只有10^6，所以可以不用二分找最靠近的数，用一个数组直接维护就可以了。

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int INF = 1e9;
const double eps = 1e-6;
const int N = 300000;
const int M = 1000010;
int cas = 1;

int a[N];
bool vis[M];
int num[M];
int n,mxnum,ans;

void run()
{
    int x;
    mxnum=0;
    memset(vis,0,sizeof(vis));
    memset(num,0,sizeof(num));
    for(int i=0;i<n;i++)
    {
        scanf("%d",&x);
        if(mxnum<x) mxnum=x;
        vis[x]=1;
    }
    for(int i=1;i<=mxnum;i++)
    {
        if(vis[i]) num[i]=i;
        else num[i]=num[i-1];
    }
    ans=0;
    for(int aj=1;aj<=mxnum;aj++)
    {
        if(!vis[aj]) continue;
        if(ans<mxnum%aj) ans=mxnum%aj;
        for(int p=aj*2;p<=mxnum;p+=aj)
        {
            if(ans < num[p-1]%aj)
                ans = num[p-1]%aj;
        }
    }
    printf("%d
",ans);
}

int main()
{
    #ifdef LOCAL
//    freopen("case.txt","r",stdin);
    #endif
    while(scanf("%d",&n)!=EOF)
        run();
    return 0;
}