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
Copy
2
input
Copy
3
1 1 1
output
Copy
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个数字，问你这些数字中能找出几对数字使得是2的幂次方
思路：题意很简单，然后思路也很好想，，暴力肯定不行，我想到用二分做，先把int范围以内2的幂都求出来，大概31位就可以了，然后2的幂减去该数组，再在原本的数组中二分查找。。。但是本题用原来的二分可能不是很好做，例如第二个样例，1 1 1，当你查找第一个的时候可能找到第一个1就结束了，但是第二个1根本没查找到，所以原来的二分可能比较难做，这里我们用到，C++的STL函数中的upper_bound和lower_bound，第一个是二分找到所有大于要查找的数，第二个二分是找到所有大于等于的数，二者再减一下就可以了。
代码：
/*
*/
#include<map>
#include<set>
#include<stack>
#include<queue>
#include<cmath>
#include<string>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
using namespace std;
#define ll unsigned long long
#define inf 0x3f3f3f
#define esp 1e-8
#define bug {printf("mmp\n");}
#define mm(a,b) memset(a,b,sizeof(a))
#define T() int test,q=1;scanf("%d",&test); while(test--)
const int maxn=2e5+100;
const double pi=acos(-1.0);
const int N=1100;
const int mod=1e9+7;
int a[maxn];
int b[maxn];
void ff()
{
a[1]=1;
for(int i=2; i<32; i++)
a[i]=a[i-1]*2;
}
int main()
{
ff();
int n;
ll cnt=0;
scanf("%d",&n);
for(int i=0; i<n; i++)
cin>>b[i];
sort(b,b+n);
for(int i=0; i<n; i++)
{
for(int j=1; j<=31; j++)
{
int ans=a[j]-b[i];
if(ans<=0)
continue;
else
{
cnt+=upper_bound(b+i+1,b+n,ans)-lower_bound(b+i+1,b+n,ans);
}
}
}
printf("%lld\n",cnt);
return 0;
}