Codeforces 702 B. Powers of Two（二分）

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B. Powers of Two

time limit per test3 seconds

memory limit per test256 megabytes

inputstandard input

outputstandard 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

4

7 3 2 1

output

2

input

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.

题目大意：

给你 n(n<=10^5) 个数a[i] (1<=i<=n)，然后，从中选两个数 a[i] ， a[j] (i < j) 满足 a[i]+a[j] = 2^x

解题思路：

因为 a[i] + a[j] 不会超过 2^31,所以 我们暴力枚举 1-31就行，然后暴力二分找 2^x - a[i] == a[j]的数，先 lower_bound()一下，然后在upper_bound一下就能找到相等的了,ans += 下标差值就行了。

My Code：

/**
2016 - 08 - 02 上午

Author: ITAK
Motto:

今日的我要超越昨日的我，明日的我要胜过今日的我，
以创作出更好的代码为目标，不断地超越自己。
**/

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <algorithm>
#include <set>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
const int MAXN = 1e6+5;
const int MOD = 1e9+7;
const double eps = 1e-7;
LL a[MAXN];
int main()
{
int n;
while(~scanf("%d",&n))
{
for(int i=0; i<n; i++)
scanf("%lld",&a[i]);
sort(a, a+n);
LL cnt = 0;
for(int i=1; i<=32; i++)
{
LL ans = (1LL<<i);
if(ans > 2*a[n-1])
break;
for(int j=0; j<n-1; j++)
{
int tp1 = lower_bound(a+j+1, a+n, ans-a[j])-a;
int tp2 = upper_bound(a+j+1, a+n, ans-a[j])-a;
if(tp1 == n)
continue;
int tmp = tp2 - tp1;
cnt += tmp;
}
}
cout<<cnt<<endl;
}
return 0;
}