cf484B

H - Maximum Value

Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Submit Status Practice CodeForces 484B

Appoint description: 
System Crawler  (2014-11-17)

Description

You are given a sequence a consisting of n integers. Find the maximum possible value of  (integer remainder of a_i divided by a_j), where 1 ≤ i, j ≤ n and a_i ≥ a_j.

Input

The first line contains integer n — the length of the sequence (1 ≤ n ≤ 2·10⁵).

The second line contains n space-separated integers a_i (1 ≤ a_i ≤ 10⁶).

Output

Print the answer to the problem.

Sample Input

Input

33 4 5

Output

2

二分真是太神了，，，枚举 一个数的倍数，最大的余数肯定是比他小，二分搜索比倍数小的数，，，，参考了网上大牛的解法，加了一个剪枝，不然会超时，无限YM

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<queue>
#include<vector>
#include<set>
using namespace std;

int a[200010];

inline int mymax(int a,int b){
    return a > b? a: b;
}

inline int bin_search(int l,int r,int val)//二分查找小于val的最大值
{
    int mid;
    while(l<=r){
        mid=(l+r)>>1;
        if(a[mid]==val) return mid-1;
        else if(a[mid]>val) r=mid-1;
        else l=mid+1;
    }
    return r;
}

int main()
{
    int n;
    while(~scanf("%d",&n)){
        for(int i=0;i<n;i++)
            scanf("%d",&a[i]);
        sort(a,a+n);
        int mmax = 0;
        for(int i=0;i<n-1;i++){
            if(a[i]!=a[i-1]){//剪枝。如果之前出现过了就不用查了；
            int tmp=a[i]+a[i];
            while(tmp<=a[n-1]){
                int pos=bin_search(0,n-1,tmp);
                mmax = mymax(mmax,a[pos]%a[i]);
                tmp+=a[i];
            }
            mmax =mymax(mmax,a[n-1]%a[i]);
            }
        }
        printf("%d\n",mmax);
    }
    return 0;
}