Codeforces Round #576 (Div. 2)
One common way of digitalizing sound is to record sound intensity at particular time moments. For each time moment intensity is recorded as a non-negative integer. Thus we can represent a sound file as an array of n non-negative integers.
If there are exactly K distinct values in the array, then we need k=⌈log2K⌉ bits to store each value. It then takes nk bits to store the whole file.
To reduce the memory consumption we need to apply some compression. One common way is to reduce the number of possible intensity values. We choose two integers l≤r, and after that all intensity values are changed in the following way: if the intensity value is within the range [l;r], we don't change it. If it is less than l, we change it to l; if it is greater than r, we change it to r. You can see that we lose some low and some high intensities.
Your task is to apply this compression in such a way that the file fits onto a disk of size I bytes, and the number of changed elements in the array is minimal possible.
We remind you that 1 byte contains 8 bits.
k=⌈log2K⌉ is the smallest integer such that K≤2k. In particular, if K=1, then k=0.
Input
The first line contains two integers n and I (1≤n≤4⋅105, 1≤I≤108) — the length of the array and the size of the disk in bytes, respectively.
The next line contains n integers ai (0≤ai≤109) — the array denoting the sound file.
Output
Print a single integer — the minimal possible number of changed elements.
Examples
input
6 1
2 1 2 3 4 3
output
2
input
6 2
2 1 2 3 4 3
output
0
input
6 1
1 1 2 2 3 3
output
2
Note
In the first example we can choose l=2,r=3. The array becomes 2 2 2 3 3 3, the number of distinct elements is K=2, and the sound file fits onto the disk. Only two values are changed.
In the second example the disk is larger, so the initial file fits it and no changes are required.
In the third example we have to change both 1s or both 3s.
1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4 #include<cmath>
5 #include<algorithm>
6 #include<map>
7 #include<set>
8 #include<vector>
9 #include<queue>
10 #include<list>
11 #include<unordered_map>
12 using namespace std;
13 #define ll long long
14 const int mod=1e9+7;
15 const ll int inf=1e18+7;
16
17 const int maxn=4e5+10;
18
19 ll int num[maxn];
20
21 int main()
22 {
23 ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
24
25 int n;
26 ll int I;
27 ll int num;
28
29 while(cin>>n>>I)
30 {
31 map<ll int,int>book;
32
33 for(int i=0;i<n;i++)
34 {
35 cin>>num;
36 book[num]++;
37 }
38
39 ll int wei=I*8/n;//题意所要限制的，最大序列中不同数的个数 的取log2
40
41 if(wei>=19)//2的20次方已经大于 maxn=4e5 ,直接特判
42 {
43 cout<<0<<endl;//空间够，不用改数字
44 continue;
45 }
46
47 wei=(1<<wei);//题意所要限制的，最大序列中不同数的个数
48
49 if(book.size()<=wei)//原来数组不同数的个数已经符合条件
50 {
51 cout<<0<<endl;//不用改数字
52 continue;
53 }
54
55 vector<int>v;
56
57 for(auto it=book.begin();it!=book.end();it++)
58 { //map对key值有序，
59 v.push_back((it->second));//将value值存入数组
60 }
61
62 ll int maxx=-1;//记录最优解
63
64 ll int now=0;
65
66 for(int i=0;i<wei;i++)//再用尺取法找最优解
67 {
68 now+=v[i];
69 }
70
71 if(now>maxx)
72 maxx=now;
73
74 for(int i=wei;i<v.size();i++)
75 {
76 now+=v[i];//往后移一位
77 now-=v[i-wei];//减去前面的
78
79 if(now>maxx)//刷新最优解
80 maxx=now;
81 }
82
83 cout<<n-maxx<<endl;//n减去最大值即为最小值
84 }
85
86 return 0;
87 }