Educational Codeforces Round 15 B
3 seconds
256 megabytes
standard input
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).
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).
Print the number of pairs of indexes i, j (i < j) that ai + aj is a power of 2.
4
7 3 2 1
2
3
1 1 1
3
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:给你n个数 问你有多少对 数的和为2^x
题解:标记每个数 枚举2^x 与a[i]取差值 统计mp[2^x-a[i]] 注意ans/2 注意开LL
/******************************
code by drizzle
blog: www.cnblogs.com/hsd-/
^ ^ ^ ^
O O
******************************/
//#include<bits/stdc++.h>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<map>
#include<algorithm>
#include<queue>
#include<cmath>
#define ll __int64
#define PI acos(-1.0)
#define mod 1000000007
using namespace std;
int n;
int a[];
map<int,int>mp;
int main()
{
scanf("%d",&n);
mp.clear();
for(int i=; i<=n; i++)
{
scanf("%d",&a[i]);
mp[a[i]]++;
}
int exm=;
ll ans=;
for(int i=; i<=; i++)
{
exm=exm<<;
for(int j=; j<=n; j++)
{
if(mp[exm-a[j]])
{
mp[a[j]]--;
ans=ans+mp[exm-a[j]];
mp[a[j]]++;
}
}
}
cout<<ans/<<endl;
return ;
}
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