Educational Codeforces Round 15 B. Powers of Two

原题链接

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 xexists
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.

从小到大枚举所有可能的2^x，然后用二分在数组中找到两个数使其和为2^x

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <queue>
#define maxn 100005
using namespace std;
typedef long long ll;

int num[maxn];
int main(){

// freopen("in.txt", "r", stdin);
int n;

scanf("%d", &n);
for(int i = 0; i < n; i++)
scanf("%d", num+i);
if(n == 1){
puts("0");
return 0;
}
sort(num, num+n);
ll s = 0;
s = num[n-2] + num[n-1];
ll ans = 0;
for(ll i = 2; i <= s; i <<= 1){
for(int j = 0; num[j] < i; j++){

ll d = i - num[j];
if(d > num[n-1] || d < num[j])
continue;
int k1 = lower_bound(num+j+1, num+n, d) - num;
int k2 = upper_bound(num+j+1, num+n, d) - num;
if(num[k1] != d)
continue;
ans += k2 - k1;
}
}
printf("%I64d\n", ans);
return 0;
}