题目链接：
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;
}