Codeforces Round #486 (Div. 3)988D. Points and Powers of Two

传送门：http://codeforces.com/contest/988/problem/D

题意：

在一堆数字中，找出尽量多的数字，使得这些数字的差都是2的指数次。

思路：

　可以知道最多有三个，差值为2、4；或者是a[i],a[i]+2的k次，a[i]-2的k次；

　　两个就直接找。

　　一个随便输出一个。

ac代码

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

#define lson (l , mid , rt << 1)
#define rson (mid + 1 , r , rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "\n";
#define pb push_back
#define pq priority_queue

#define Pll pair<ll,ll>
#define Pii pair<int,int>

#define fi first
#define se second

#define OKC ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
typedef long long ll;
typedef unsigned long long ull;

/*-----------------show time----------------*/
const int maxn = 2e5+;
int n;
ll a[maxn];
map<ll,int>mp;
ll ksm (int a,int b)
{
    ll res = 1ll;
    while(b)
    {
        if(b&)res = res * a;
        a = a*a;
        b>>=;
    }
    return res;
}
int main(){
    scanf("%d", &n);
    // cout<<ksm(2,1)<<endl;
    for(int i=; i<=n; i++)
    {
        scanf("%lld" , &a[i]);
        if(!mp.count(a[i]))mp[a[i]] = ;
        mp[a[i]]++;
    }
    int flag2 = ,flag1 = ;
    ll ans,jie;
    for(int i=; i<=n&&flag1==; i++)
    {
        if(mp.count(a[i] + )&&mp.count(a[i]+))
        {
            flag2 = ;
            ans = a[i];
            break;
        }
        for(int j=; j<; j++)
        {
            if(mp.count(a[i]+ksm(,j))&&mp.count(a[i]-ksm(,j)))
            {
                ans = a[i];
                jie = j;
                flag1 = ;
                break;
            }
        }
    }
    if(flag2)
    {
        printf("3\n%lld %lld %lld\n",ans,ans+,ans+);
        return ;
    }
    if(flag1)
    {
        printf("3\n%lld %lld %lld\n",ans,ans+ksm(,jie),ans-ksm(,jie));
        return ;
    }
    for(int i=; i<=n && flag2 == ; i++)
    {
        for(int j=; j<; j++)
        {
            if(mp.count(a[i]+ksm(,j)))
            {
                ans = a[i];
                jie = j;
                flag2 = ;
                break;
            }
        }
    }
    if(flag2)
    {
        printf("2\n%lld %lld\n",ans,ans+ksm(,jie));
        return ;
    }
    printf("1\n%lld\n",a[]);

}

CF988D