Codeforces 702B - Powers of Two

最新推荐文章于 2021-05-20 10:11:26 发布

nare123

最新推荐文章于 2021-05-20 10:11:26 发布

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分类专栏：二分 map

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本文介绍了一种算法，用于找出整数数组中所有满足两数之和为2的幂的数对。通过排序数组并使用二分查找提高效率，复杂度为O(nlog(n)log(m))。示例代码展示了具体的实现细节。

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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 
   
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.

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 long
map<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);
}