Codeforces 702 B Powers of Two

B. Powers of Two

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given n integers a1, a2, ..., an. Find the number of pairs of indexes i, j (i < j) that ai + aj is a power of 2 (i. e. some integer x exists so that ai + aj = 2x).

Input

The first line contains the single positive integer n (1 ≤ n ≤ 105) — the number of integers.

The second line contains n positive integers a1, a2, ..., an (1 ≤ ai ≤ 109).

Output

Print the number of pairs of indexes i, j (i < j) that ai + aj is a power of 2.

Examples

input

Copy

4
7 3 2 1

output

2

input

Copy

3
1 1 1

output

3

Note

In the first example the following pairs of indexes include in answer: (1, 4) and (2, 4).

In the second example all pairs of indexes (i, j) (where i < j) include in answer.

【题意】

似乎很好理解

【思路】

map  数组标记一下就可以 

注意 如果打表的话 会超时，   扫描1-64 也会超时

用 0-32   << 做一操作

【代码实现】

//#include <bits/stdc++.h>
#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <cmath>
#include <math.h>
#include <cstring>
#include <string>
#include <queue>
#include <deque>
#include <stack>
#include <stdlib.h>
#include <list>
#include <map>
#include <utility>
#include <set>
#include <bitset>
#include <vector>
#define mem(a,b) memset(a,b,sizeof(a))
#define findx(x,b,n) lower_bound(b+1,b+1+n,x)-b
#define FIN      freopen("input.txt","r",stdin)
#define FOUT     freopen("output.txt","w",stdout)
#define SHUT ios_base::sync_with_stdio(false); cout.setf(ios::fixed); cout.precision(20); cout.tie(nullptr); cin.tie(nullptr);
#define lson rt << 1, l, mid
#define rson rt << 1|1, mid + 1, r
#define  FI(n) IO::read(n)
#define  Be IO::begin()

using namespace std;
typedef long long ll;
const double PI=acos(-1);
const int INF=0x3f3f3f3f;
const double esp=1e-6;
const int maxn=1e6+5;
const int MAXN=50005;
const int MOD=1e9+7;
const int mod=1e9+7;
int dir[5][2]={0,1,0,-1,1,0,-1,0};
/*
namespace IO {
	const int MT = 5e7;
	char buf[MT]; int c,sz;
	void begin(){
		c = 0;
		sz = fread(buf, 1, MT, stdin);//一次性输入
	}
	template<class T>
	inline bool read(T &t){
		while( c < sz && buf[c] != '-' && ( buf[c]<'0' || buf[c] >'9')) c++;
		if( c>=sz) return false;
		bool flag = 0; if( buf[c]== '-') flag = 1,c++;
		for( t=0; c<=sz && '0' <=buf[c] && buf[c] <= '9'; c++ ) t= t*10 + buf[c]-'0';
		if(flag) t=-t;
		return true;
	}
}
*/
/*
inline void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){if(!b){ x=1; y=0; d=a; }else{ ex_gcd(b,a%b,d,y,x); y-=x*(a/b);};}
inline ll gcd(ll a,ll b){ return b?gcd(b,a%b):a;}
inline ll exgcd(ll a,ll b,ll &x,ll &y){if(!b){x=1;y=0;return a;}ll ans=exgcd(b,a%b,x,y);ll temp=x;x=y;y=temp-a/b*y;return ans;}
inline ll lcm(ll a,ll b){ return b/gcd(a,b)*a;}

inline ll inv_exgcd(ll a,ll n){ll d,x,y;ex_gcd(a,n,d,x,y);return d==1?(x+n)%n:-1;}
inline ll inv1(ll b){return b==1?1:(MOD-MOD/b)*inv1(MOD%b)%MOD;}
inline ll inv2(ll b){return qpow(b,MOD-2);}*/
inline ll qpow(ll x,ll n){ll res=1;for(;n;n>>=1){if(n&1)res=(res*x);x=(x*x);}return res;}
map<ll,ll>mp,vis;

int main()
{
    SHUT;
    //init();
    ll n;
    cin>>n;
    ll ans=0;
    ll x;
    for(int i=1;i<=n;i++)
    {
        cin>>x;
        for(int j=0;j<32;j++)
        {
            ans+=mp[(1<<j)-x];
        }
        mp[x]++;

    }
    cout<<ans<<endl;
    return 0;
}

123