思路:ai+aj=2^x,移过去就是aj=2^x-ai嘛,由于ai的范围到10^9,所以其实x不会超过31,所以枚举一下x就可以了
#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5+7;
#define LL long long
map<LL,int>vis;
LL ans = 0;
int main()
{
int n=0;
scanf("%d",&n);
for(int i = 1;i<=n;i++)
{
LL t;scanf("%lld",&t);
for(int j = 0;j<=31;j++)
ans+=vis[(1LL<<j)-t];
vis[t]++;
}
printf("%lld\n",ans);
}
B. Powers of Two
time limit per test
memory limit per test
input
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.