CodeForce 484B：（最大余数）

a[i]=k*a[j]+m

m即是所求

从大到小枚举a[j]后，枚举倍数k

用到了stl里的low_bound()

#include"cstdio"
#include"cstring"
#include"algorithm"
#define MAXN 1000005
using namespace std;
int num[MAXN],n;
int get_ans(int x)
{   int temp=x,ans=0;
    while(temp<num[n-1]){
        temp+=x;
        int pos=lower_bound(num,num+n,temp)-num;
        if(!pos) continue;
        else pos--;
        ans=max(ans,num[pos]%x);
    }
    return ans;
}
int main()
{   while(scanf("%d",&n)!=EOF){
        for(int i=0;i<n;i++)
            scanf("%d",&num[i]);
        sort(num,num+n);
        int ans=0;
        for(int i=n-1;i>=0;i--){
            if(ans>=num[i]) break;
            if(num[i]==num[i+1]&&i!=n-1) continue;
            ans=max(ans,get_ans(num[i]));
        }
        printf("%d
",ans);
    }
    return 0;
}