codeforce日记 371A

题目：

A. K-Periodic Array

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

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 

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.

Sample test(s)

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.

解题思路：

对于每个周期为k的数组来说 a1,=a（1+i*k）。所以对这样的序列来说记录2和1的个数，其差值就是需要改变的最小数目。依次类推得到a2……ak需要改变的最小数目，求和即可。

AC代码：

#include <iostream>;

using namespace std;

int main(){
int n,k;
int a[100];
int numof2=0,numof1=0;
int change=0;
cin>>n>>k;
for(int i=0;i<n;i++){
cin>>a[i];
}
for(int i=0;i<k;i++){
for(int j=0;j<n/k;j++){
if(a[i+j*k]==1){
numof1++;
}
else
{
numof2++;
}
}
change+=min(numof2,numof1);
numof1=0;
numof2=0;
}
cout<<change;

}