Codeforces 702B - 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 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

47 3 2 1

output

2

input

31 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：让你找有多少对儿ai+aj=2^x (x属于lg(a1+a2)~lg(an-1+an)假设a数组是从小到大排列的)

先排下序，然后对j(1~n)从i+1~n找有没有aj满足条件，这个过程可以用二分来实现。复杂度O(nlg(n)*lg(m))吧（不太确定）。

（这段代码自己感觉写的好矬。。）

#include<stdio.h>#include<iostream>#include<algorithm>#include<string>#include<string.h>#include<map>using namespace std;int a[101000];#define ll long longmap<int,int>mapp;map<int,int>use;int lg2(ll x){    ll s=1;    int cnt=0;    while (s*2<=x)    {        s*=2;        cnt++;    }    return cnt;}int _2n(int x){    int s=1;    for (int i=1;i<=x;i++)    {        s*=2;    }    return s;}int main(){    int n;    scanf("%d",&n);    mapp.clear();    use.clear();    for (int i=1;i<=n;i++) {scanf("%d",&a[i]); mapp[a[i]]++;}    sort(a+1,a+n+1); int rl=lg2(a[1]+a[2]);int rr=lg2(a[n]+a[n-1]);    ll ans=0;    for (int i=1;i<=n;i++)    {       for (int j=rl;j<=rr;j++)       {           int l=i;int r=n; int ta=_2n(j);           int fi=ta-a[i];           if (fi>0)           {             while (r-l>1)             {                 int mid=(l+r)/2;                 if (fi<=a[mid]) r=mid;                 else l=mid;             }           if (a[i]+a[r]==ta) {               if (a[i]==a[r])               {ans=ans+mapp[a[r]]-use[a[r]]-1;use[a[r]]++;}               else {ans=ans+mapp[a[r]];}            }           }       }    }    printf("%I64d",ans);}