codeforces 484B B. Maximum Value(二分)

题目链接：

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.

Examples

input

3
3 4 5

output

2

题意：

给n个数,要求算出最大的a[i]%a[j]的值,其中a[i]>=a[j];

思路：

可以枚举a[j],然后找出它的所有倍数,然后再在序列中找到小于它且最接近它的数.然后就是这个阶段里面最大的值了;

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <bits/stdc++.h>
#include <stack>
#include <map>

using namespace std;

#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));

typedef  long long LL;

template<class T> void read(T&num) {
    char CH; bool F=false;
    for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
    for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
    F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
    if(!p) { puts("0"); return; }
    while(p) stk[++ tp] = p%10, p/=10;
    while(tp) putchar(stk[tp--] + '0');
    putchar('\n');
}

const LL mod=1e9+7;
const double PI=acos(-1.0);
const int inf=1e9;
const int N=2e5+20;
const int maxn=1e6+220;
const double eps=1e-12;

int a[N],n,vis[maxn];

int main()
{
    read(n);
    For(i,1,n)read(a[i]);
    sort(a+1,a+n+1);
    int ans=0;
    for(int i=1;i<=n;i++)
    {
        if(a[i]==a[i-1])continue;
        int pos=i;
        for(int k=2; ;k++)
        {
            int h=k*a[i];
            int l=pos+1,r=n;
            while(l<=r)
            {
                int mid=(l+r)>>1;
                if(a[mid]>=h)r=mid-1;
                else l=mid+1;
            }
            pos=r;
            ans=max(ans,a[pos]%a[i]);
            if(h>=maxn)break;
        }
    }
    cout<<ans<<endl;
    return 0;
}