A. K-Periodic Array----思维

最新推荐文章于 2022-09-14 09:59:42 发布

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最新推荐文章于 2022-09-14 09:59:42 发布

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本文探讨了一道编程题，目标是最小化改变数组元素数量，使数组成为k周期。通过将数组拆分为k组并计算每组中1和2的数量差异来实现。

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This task will exclusively concentrate only on the arrays where all elements equal 1 and/or 2.

Array a is k-period if its length is divisible by k and there is such array b of length k, that a is represented by array b written exactly $\text{[math]}$ times consecutively. In other words, array a is k-periodic, if it has period of length k.

For example, any array is n-periodic, where n is the array length. Array [2, 1, 2, 1, 2, 1] is at the same time 2-periodic and 6-periodic and array [1, 2, 1, 1, 2, 1, 1, 2, 1] is at the same time 3-periodic and 9-periodic.

For the given array a, consisting only of numbers one and two, find the minimum number of elements to change to make the array k-periodic. If the array already is k-periodic, then the required value equals 0.

Input

The first line of the input contains a pair of integers n, k (1 ≤ k ≤ n ≤ 100), where n is the length of the array and the value n is divisible by k. The second line contains the sequence of elements of the given array a1, a2, ..., an (1 ≤ ai ≤ 2), ai is the i-th element of the array.

Output

Print the minimum number of array elements we need to change to make the array k-periodic. If the array already is k-periodic, then print0.

Examples

input
6 2
2 1 2 2 2 1

output
1

input
8 4
1 1 2 1 1 1 2 1

output
0

input
9 3
2 1 1 1 2 1 1 1 2

output
3

Note

In the first sample it is enough to change the fourth element from 2 to 1, then the array changes to [2, 1, 2, 1, 2, 1].

In the second sample, the given array already is 4-periodic.

In the third sample it is enough to replace each occurrence of number two by number one. In this case the array will look as[1, 1, 1, 1, 1, 1, 1, 1, 1] — this array is simultaneously 1-, 3- and 9-periodic.

题目链接：http://codeforces.com/contest/371/problem/A

题目的意思是说给你n个数，只有1和2，现在k个为一组，问你最少改变几个数，可以使k组里面的数都相同。

思路很简单，我们把k组排成一个二维数组，只需要把答案加上每一列1和2比较少的个数即可。

代码：

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
int a[10000];
int main(){
    int n,k;
    scanf("%d%d",&n,&k);
    for(int i=0;i<n;i++){
        scanf("%d",&a[i]);
    }
    int ans=0;
    for(int i=0;i<k;i++){
        int cnt=0;
        for(int j=i;j<n;j+=k){
            if(a[j]==1)
                cnt++;
        }
        ans+=min(cnt,n/k-cnt);
    }
    cout<<ans<<endl;
    return 0;
}