Codeforces 1005C（贪心）

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原文链接：http://www.cnblogs.com/Chen-Jr/p/11007270.html

本文针对CF C题提供了解决方案，介绍了如何通过贪心算法和使用map数据结构来判断序列中是否存在两个不同的元素，其和为2的幂，并给出了具体的实现代码。

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题面：

A sequence $a_{1}, a_{2}, \dots, a_{n}$ is called good if, for each element $a_{i}$ , there exists an element $a_{j}$ ( $i \neq j$ ) such that $a_{i} + a_{j}$ is a power of two (that is, $2^{d}$ for some non-negative integer $d$ ).

For example, the following sequences are good:

$[5, 3, 11]$ (for example, for $a_{1} = 5$ we can choose $a_{2} = 3$ . Note that their sum is a power of two. Similarly, such an element can be found for $a_{2}$ and $a_{3}$ ),
$[1, 1, 1, 1023]$ ,
$[7, 39, 89, 25, 89]$ ,
$[]$ .

Note that, by definition, an empty sequence (with a length of $0$ ) is good.

For example, the following sequences are not good:

$[16]$ (for $a_{1} = 16$ , it is impossible to find another element $a_{j}$ such that their sum is a power of two),
$[4, 16]$ (for $a_{1} = 4$ , it is impossible to find another element $a_{j}$ such that their sum is a power of two),
$[1, 3, 2, 8, 8, 8]$ (for $a_{3} = 2$ , it is impossible to find another element $a_{j}$ such that their sum is a power of two).

You are given a sequence $a_{1}, a_{2}, \dots, a_{n}$ . What is the minimum number of elements you need to remove to make it good? You can delete an arbitrary set of elements.

Input

The first line contains the integer $n$ ( $1 \leq n \leq 120000$ ) — the length of the given sequence.

The second line contains the sequence of integers $a_{1}, a_{2}, \dots, a_{n}$ ( $1 \leq a_{i} \leq 10^{9}$ ).

Output

Print the minimum number of elements needed to be removed from the given sequence in order to make it good. It is possible that you need to delete all $n$ elements, make it empty, and thus get a good sequence.

   Examples
  
     input
     
      Copy
     
6
4 7 1 5 4 9

     output
     
      Copy
     
1

     input
     
      Copy
     
5
1 2 3 4 5

     output
     
      Copy
     
2

     input
     
      Copy
     
1
16

     output
     
      Copy
     
1

     input
     
      Copy
     
4
1 1 1 1023

     output
     
      Copy
     
0

Note

In the first example, it is enough to delete one element $a_{4} = 5$ . The remaining elements form the sequence $[4, 7, 1, 4, 9]$ , which is good.

题目描述：

给你一个大小为n的数组，问你对于每个数ai，使得ai+aj都为2的幂，问你不符合条件的个数。

题目分析：

事实上这个题跟之前的CF1003D很相似。因为要求我们判断是否和为2的幂，因此我们可以贪心的从2^30次方往下进行枚举，如果数组中的每一个数ai，在数组中存在j-ai的话，则证明符合条件，反之则不符合。

而因为题目的数据范围较大，因此我们需要开map来进行存储。

值得注意的是，因为题目要保证ai与bi不相同，因此得小心处理好ai与bi相同的情况，防止多解。

代码：

#include <bits/stdc++.h>
#define maxn 120005
using namespace std;
typedef long long ll;
map<ll,ll>mp;
map<ll,bool>vis;
ll a[maxn];
int main()
{
    int n;
    cin>>n;
    for(int i=0;i<n;i++){
        scanf("%I64d",&a[i]);
        mp[a[i]]++;
    }
    sort(a,a+n);
    int cnt=0;
    for(int i=0;i<n;i++){
        if(vis[a[i]]) continue;
        bool flag=true;
        for(ll j=1<<30;j>=1;j>>=1){
            if(j<a[i]) break;
            if(mp.count(j-a[i])){//如果在map中存在j-a[i]，则证明原数列中存在j-a[i]，则a[i]符合条件
                if(j-a[i]==a[i]&&mp[j-a[i]]==1) continue;//如果j-a[i]与a[i]相等，同时map中a[i]只有一个，则证明不能符合条件
                vis[a[i]]=1;
                vis[j-a[i]]=1;
                flag=false;
                break;
            }
        }
        if(flag) cnt++;//不符合则答案+1
    }
    cout<<cnt<<endl;
}

转载于:https://www.cnblogs.com/Chen-Jr/p/11007270.html