Summarize to the Power of Two（map+思维）

Posted 2020-12-27 wkfvawl

A sequence ${\mathrm{a1,a2,\dots ,ana1,a2,\dots ,an is\; called\; good\; if,\; for\; each\; element aiai,\; there\; exists\; an\; element ajaj (i\ne ji\ne j)\; such\; that ai+ajai+aj is\; a\; power\; of\; two\; (that\; is, 2d2d for\; some\; non-negative\; integer dd).}}^{}$

For example, the following sequences are good:

• $[5,3,11][5,3,11] (for\; example,\; for a1=5a1=5 we\; can\; choose a2=3a2=3.\; Note\; that\; their\; sum\; is\; a\; power\; of\; two.\; Similarly,\; such\; an\; element\; can\; be\; found\; for a2a2 and a3a3),$
• $[1,1,1,1023][1,1,1,1023],$
• $[7,39,89,25,89][7,39,89,25,89],$
• $[][].$

Note that, by definition, an empty sequence (with a length of $\mathrm{00)\; is\; good.}$

For example, the following sequences are not good:

• $[16][16] (for a1=16a1=16,\; it\; is\; impossible\; to\; find\; another\; element ajaj such\; that\; their\; sum\; is\; a\; power\; of\; two),$
• $[4,16][4,16] (for a1=4a1=4,\; it\; is\; impossible\; to\; find\; another\; element ajaj such\; that\; their\; sum\; is\; a\; power\; of\; two),$
• $[1,3,2,8,8,8][1,3,2,8,8,8] (for a3=2a3=2,\; it\; is\; impossible\; to\; find\; another\; element ajaj such\; that\; their\; sum\; is\; a\; power\; of\; two).$

You are given a sequence ${\mathrm{a1,a2,\dots ,ana1,a2,\dots ,an.\; What\; is\; the\; minimum\; number\; of\; elements\; you\; need\; to\; remove\; to\; make\; it\; good?\; You\; can\; delete\; an\; arbitrary\; set\; of\; elements.}}^{}$

Input

The first line contains the integer $\mathrm{nn (1\le n\le 1200001\le n\le 120000)\; —\; the\; length\; of\; the\; given\; sequence.}$

The second line contains the sequence of integers ${\mathrm{a1,a2,\dots ,ana1,a2,\dots ,an (1\le ai\le 1091\le ai\le 109).}}^{}$

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 $\mathrm{nn elements,\; make\; it\; empty,\; and\; thus\; get\; a\; good\; sequence.}$

Examples

input

6
4 7 1 5 4 9

output

1

input

5
1 2 3 4 5

output

Copy

2

input

1
16

output

1

input

4
1 1 1 1023

output

0

Note

In he first example, it is enough to delete one element ${\mathrm{a4=5.\; The\; remaining\; elements\; form\; the\; sequence [4,7,1,4,9][4,7,1,4,9],\; which\; is\; good.}}^{}$

 

题目意思：给你一个数列，然后让你删掉某个数，让每一个数都可以与另外一个数相加，之和等于2的^d次方，问最少删掉的个数

解题思路：先预处理出2的次幂，然后暴力枚举出每个数与每个2的次幂之差，去判断在map中是否有这样的差存在（注意相同的两个数，比如4 ，4   这个时候就可以特判一下，出现次数），若不存在那么就需要删除掉这个数。

 

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<map>
 4 #include<algorithm>
 5 #define ll long long int
 6 using namespace std;
 7 ll d[32];
 8 ll a[1000010];
 9 map<ll,ll>mp;
10 void double_2()
11 {
12     int i;
13     d[0]=1;
14     for(i=1; i<=31; i++)
15     {
16         d[i]=d[i-1]*2;
17     }
18 }
19 int main()
20 {
21     int i,j,flag,n;
22     ll k;
23     double_2();///2的幂
24     while(scanf("%d",&n)!=EOF)
25     {
26         int count=0;
27         mp.clear();///map清零
28         for(i=0; i<n; i++)
29         {
30             scanf("%lld",&a[i]);
31             mp[a[i]]++;///记录出现的次数
32         }
33         for(i=0; i<n; i++)
34         {
35             flag=0;
36             for(j=0; j<=31; j++)
37             {
38                 if(a[i]>=d[j])
39                 {
40                     continue;
41                 }
42                 else
43                 {
44                     k=d[j]-a[i];
45                     if(mp[k])///map中存在
46                     {
47                         if(a[i]==k)
48                         {
49                             if(mp[k]>=2)///map中需要至少两个
50                             {
51                                 flag=1;
52                                 break;
53                             }
54                         }
55                         else
56                         {
57                             flag=1;
58                             break;
59                         }
60                     }
61                 }
62             }
63             if(!flag)
64             {
65                 count++;
66             }
67         }
68         printf("%d
",count);
69     }
70     return 0;
71 }