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.
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.
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.
6 2 2 1 2 2 2 1
1
8 4 1 1 2 1 1 1 2 1
0
9 3 2 1 1 1 2 1 1 1 2
3
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
#include
#include
using namespace std;
int a[10000];
int main(){
int n,k;
scanf("%d%d",&n,&k);
for(int i=0;i