Word s of length n is called k-complete if
s is a palindrome, i.e. si=sn+1−i for all 1≤i≤n;
s has a period of k, i.e. si=sk+i for all 1≤i≤n−k.
For example, “abaaba” is a 3-complete word, while “abccba” is not.
Bob is given a word s of length n consisting of only lowercase Latin letters and an integer k, such that n is divisible by k. He wants to convert s to any k-complete word.
To do this Bob can choose some i (1≤i≤n) and replace the letter at position i with some other lowercase Latin letter.
So now Bob wants to know the minimum number of letters he has to replace to convert s to any k-complete word.
Note that Bob can do zero changes if the word s is already k-complete.
You are required to answer t test cases independently.
Input
The first line contains a single integer t (1≤t≤105) — the number of test cases.
The first line of each test case contains two integers n and k (1≤k<n≤2⋅105, n is divisible by k).
The second line of each test case contains a word s of length n.
It is guaranteed that word s only contains lowercase Latin letters. And it is guaranteed that the sum of n over all test cases will not exceed 2⋅105.
Output
For each test case, output one integer, representing the minimum number of characters he has to replace to convert s to any k-complete word.
Example
Input
4
6 2
abaaba
6 3
abaaba
36 9
hippopotomonstrosesquippedaliophobia
21 7
wudixiaoxingxingheclp
Output
2
0
23
16
Note
In the first test case, one optimal solution is aaaaaa.
In the second test case, the given word itself is k-complete.
唉，挺简单的一道题目给整复杂了。。
思路：根据题目要求，我们可以推断出，给定的字符串，每k位就是一个回文字符串。这样的话，我们就考虑这k位的字符串。对于这k位的回文字符串，我们要考虑对称的位置字符的数量，例如第一个样例来说，abaaba，第一位a有2个，b有1个；第二位a有2个，b有1个。因为第一位和第二位是对称的，所以我们要合起来考虑，a一共有4个，b一共有2个，a的数量更多，因此这些位置要替换成a。思路就是这样，实现的方式有很多。但是不要盲目的memset，因为有可能超时。
代码如下：
#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int maxx=2e5+100;
int cnt[maxx][27];
string s;
int n,k;
inline void init()
{
for(int i=0;i<k;i++) memset(cnt[i],0,sizeof(cnt[i]));
}
inline void dfs(int u,int flag)
{
if(u>=n) return ;
cnt[flag][s[u]-'a']++;
dfs(u+k,flag);
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&k);
cin>>s;
init();
for(int i=0;i<k;i++) dfs(i,i);
int l=0,r=k-1;
int ans=0,_max=0;
while(l<=r)//这样写就不用把奇数的情况单独拿出来考虑了。
{
if(l==r)
{
_max=0;
for(int i=0;i<26;i++) _max=max(_max,cnt[l][i]);
ans+=_max;
break;
}
else
{
_max=0;
for(int i=0;i<26;i++) _max=max(_max,cnt[l][i]+cnt[r][i]);
ans+=_max;
l++,r--;
}
}
cout<<n-ans<<endl;
}
return 0;
}
努力加油a啊，(o)/~
