CodeForces 702B Powers of Two

简单题。

开一个$map$记录一下每个数字出现了几次，那么读入的时候$f[a[i]]+1$。

计算$a[i]$做出的贡献的时候，先把$f[a[i]]-1$，然后再枚举$x$，答案加上$f[{2^x} - a[i]]$。

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<iostream>
using namespace std;
typedef long long LL;
const double pi=acos(-1.0),eps=1e-8;
void File()
{
    freopen("D:\in.txt","r",stdin);
    freopen("D:\out.txt","w",stdout);
}
template <class T>
inline void read(T &x)
{
    char c = getchar(); x = 0;while(!isdigit(c)) c = getchar();
    while(isdigit(c)) { x = x * 10 + c - '0'; c = getchar();  }
}

const int maxn=100010;
LL a[maxn],ans;
map<LL,int>f;
int n;

int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++) scanf("%lld",&a[i]),f[a[i]]++;
    for(int i=1;i<=n;i++)
    {
        f[a[i]]--; LL sum=1;
        for(int j=0;j<=32;j++)
        {
            ans=ans+f[sum-a[i]];
            sum=sum*2;
        }
    }
    printf("%lldn",ans);
    return 0;
}