A. MP3
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
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
inputCopy
6 1
2 1 2 3 4 3
outputCopy
2
inputCopy
6 2
2 1 2 3 4 3
outputCopy
0
inputCopy
6 1
1 1 2 2 3 3
outputCopy
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.
尺取法的应用
将该题转化为连续序列不同元素个数为K的最长序列
对原数组进行排序，变成连续序列，尺取区间检查序列中不同元素个数为K的解得到最优解
#include <cstdio>
#include <algorithm>
using namespace std;
const int MAXN = 4e5 + 10;
int a[MAXN], b[MAXN];
int main()
{
int n, I;
scanf("%d%d", &n, &I);
I = I * 8 / n;
//求出数组中最大不同元素的个数K
int K = I > 30 ? 1 << 30 : 1 << I;
for (int i = 0; i < n; ++i)
{
scanf("%d", &a[i]);
}
sort(a, a+n);
//用数组b来存储a中对应区间的不同元素个数，b[i]表示a数组[0,i]区间不同元素的个数
b[0] = 1;
for (int i = 1; i < n; ++i)
{
if (a[i] == a[i - 1])
{
b[i] = b[i - 1];
}
else
{
b[i] = b[i - 1] + 1;
}
}
//如果原数组中不同元素的个数小于等于K，那么不用压缩
if (K >= b[n - 1])
{
printf("0\n");
return 0;
}
int l, r, minlen = n, ans;
//尺取法，满足条件是数组中不同元素个数等于K
//让ans不断变大，也就是所含区间不同元素个数变多直到大于等于K
//当ans等于K，就是一个解；ans大于K，缩短左区间，减少元素个数，使ans变小
for (l = r = 0; r < n; ++r)
{
//延伸右区间
ans = b[r];
//ans>=K才可能有解
if (ans < K)
continue;
//缩小左区间
while (ans - b[l] >= K)
ans -= b[l++];
//筛选答案
if (ans == K)
{
minlen = min(minlen, n - (r - l + 1));
}
}
printf("%d\n", minlen);
return 0;
}
