【題目描述】
The only difference between easy and hard versions is the size of the input.

You are given a string s
consisting of n

characters, each character is ‘R’, ‘G’ or ‘B’.

You are also given an integer k
. Your task is to change the minimum number of characters in the initial string s so that after the changes there will be a string of length k that is a substring of s

, and is also a substring of the infinite string “RGBRGBRGB …”.

A string a
is a substring of string b if there exists a positive integer i such that a1=bi, a2=bi+1, a3=bi+2, …, a|a|=bi+|a|?1

. For example, strings “GBRG”, “B”, “BR” are substrings of the infinite string “RGBRGBRGB …” while “GR”, “RGR” and “GGG” are not.

You have to answer q

independent queries.

Input

The first line of the input contains one integer q

(1≤q≤2?105) — the number of queries. Then q

queries follow.

The first line of the query contains two integers n
and k (1≤k≤n≤2?105) — the length of the string s

and the length of the substring.

The second line of the query contains a string s
consisting of n

characters ‘R’, ‘G’ and ‘B’.

It is guaranteed that the sum of n
over all queries does not exceed 2?105 (∑n≤2?105

).

Output

For each query print one integer — the minimum number of characters you need to change in the initial string s

so that after changing there will be a substring of length k in s

that is also a substring of the infinite string "RGBRGBRGB ...".

Example
Input

3 5 2 BGGGG 5 3 RBRGR 5 5 BBBRROutput1 0 3

【題目分析】
題目要求給定串改變最少的字符使串的一個子串變成一個無限長串$"RGBRGBRGB\dots \dots "$的一個子串
我自己看完后十分懵逼，因為沒什么想法，既沒有說改變的是哪一部分子串，也沒有說變成什么樣子，還需要最小。在看到別人的博客以后才恍然大悟，我們不必要將目光局限在改變的那一小部分子串上，我們?nèi)绻麑⒆罱K改變的子串拓展成和給定串長度相同的串，無外乎三種，分別以$R$、$G$、$B$開頭，我們就分別計算如果將串變成那三種樣子哪些位置需要改變（需要改變的位置記為1，不需要改變的位置記為0，然后用一個前綴和數(shù)組進(jìn)行維護(hù)），然后再找長度為k的子串中需要改變最少的。
覺得自己這種抽象能力不太夠，還需要多進(jìn)行鍛煉。
【AC代碼】

#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include<vector> #include<queue> #include<set> #include<climits> #include<cstdlib> #include<cmath>using namespace std;typedef long long ll;const int MAXN=200005; char s[MAXN]; const char t[4]="RGB"; int sum[MAXN][3]; int ans;int main() {int T,n,k;scanf("%d",&T);while(T--){scanf("%d%d",&n,&k);scanf("%s",s);ans=INT_MAX;//memset(sum,0,sizeof(sum));sum[0][0]=sum[0][1]=sum[0][2]=0;for(int i=0;i<n;i++){for(int j=0;j<3;j++){if(s[i]!=t[(i+j)%3]){sum[i+1][j]=sum[i][j]+1;}else{sum[i+1][j]=sum[i][j];}}}for(int i=0;i<=n-k;i++){for(int j=0;j<3;j++){ans=min(ans,sum[i+k][j]-sum[i][j]);}}printf("%d\n",ans);}return 0; }

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