D1. RGB Substring (easy version)
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
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≤2000) — the number of queries. Then q queries follow.
The first line of the query contains two integers n and k (1≤k≤n≤2000) — 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 2000 (∑n≤2000).
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
inputCopy
3
5 2
BGGGG
5 3
RBRGR
5 5
BBBRR
outputCopy
1
0
3
Note
In the first example, you can change the first character to 'R' and obtain the substring "RG", or change the second character to 'R' and obtain "BR", or change the third, fourth or fifth character to 'B' and obtain "GB".
In the second example, the substring is "BRG".
题意：
两题只是数据范围不同，所以放在一起讲解，给你一个字符串和一个k，问你最少改变多少个字符可以使字符串的一个k长度子串是RGBRGBRGB***** 无限长的子串？
思路：
D1： 时间复杂度显然可以O(n*k) 来做
我们先构造3个k长度的RGB字符串，分别以 R，G，B 开头。
然后去O(nk)枚举字符串的所有k长度的连续子串O(3K)的扫描这个子串和3个GBR字符串的不同字符个数，
维护最小的不同个数就是题目答案。
D2 : 时间复杂度O(N+K)是可以接受的。
那么我们先构造3个长度为n的RGB字符串，让每一个都和字符串str继续匹配。
匹配时我们维护a[i] 代表第i个字符串是否与RGB字符串不同，不同为1，相同为0.
然后求a[i]的前缀和，然后Ｏ（Ｎ）扫一遍处理长度为k的字符串最少有多少个与RGB字符串的不同值。
两份代码：第一个是D1，第二个是D2
细节见代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define rt return
#define dll(x) scanf("%I64d",&x)
#define xll(x) printf("%I64d\n",x)
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define chu(x) cout<<"["<<#x<<" "<<(x)<<"]"<<endl
using namespace std;
typedef long long ll;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a,ll b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
inline void getInt(int* p);
const int maxn=1000010;
const int inf=0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
char s[maxn];
int main()
{
//freopen("D:\\common_text\\code_stream\\in.txt","r",stdin);
//freopen("D:\\common_text\code_stream\\out.txt","w",stdout);
int q;
int n,k;
gbtb;
cin>>q;
while(q--)
{
cin>>n>>k;
cin>>s;
string str1,str2,str3;
str1="";
str2="";
str3="";
for(int i=0;i<k;++i)
{
if(i%3==0)
{
str1.pb('R');
}else if((i%3)==1)
{
str1.pb('G');
}
else
{
str1.pb('B');
}
}
for(int i=0;i<k;++i)
{
if(i%3==0)
{
str2.pb('G');
}else if((i%3)==1)
{
str2.pb('B');
}
else
{
str2.pb('R');
}
}
for(int i=0;i<k;++i)
{
if(i%3==0)
{
str3.pb('B');
}else if((i%3)==1)
{
str3.pb('R');
}
else
{
str3.pb('G');
}
}
int ans=inf;
for(int i=0;i+k-1<n;++i)
{
int cnt=0;
for(int j=i;j<=i+k-1;++j)
{
if(s[j]!=str1[j-i])
{
cnt++;
}
}
ans=min(ans,cnt);
cnt=0;
for(int j=i;j<=i+k-1;++j)
{
if(s[j]!=str2[j-i])
{
cnt++;
}
}
ans=min(ans,cnt);
cnt=0;
for(int j=i;j<=i+k-1;++j)
{
if(s[j]!=str3[j-i])
{
cnt++;
}
}
ans=min(ans,cnt);
cnt=0;
}
cout<<ans<<endl;
}
return 0;
}
inline void getInt(int* p) {
char ch;
do {
ch = getchar();
} while (ch == ' ' || ch == '\n');
if (ch == '-') {
*p = -(getchar() - '0');
while ((ch = getchar()) >= '0' && ch <= '9') {
*p = *p * 10 - ch + '0';
}
}
else {
*p = ch - '0';
while ((ch = getchar()) >= '0' && ch <= '9') {
*p = *p * 10 + ch - '0';
}
}
}
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define rt return
#define dll(x) scanf("%I64d",&x)
#define xll(x) printf("%I64d\n",x)
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define chu(x) cout<<"["<<#x<<" "<<(x)<<"]"<<endl
using namespace std;
typedef long long ll;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a,ll b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
inline void getInt(int* p);
const int maxn=1000010;
const int inf=0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
char s[maxn];
int a[maxn];
int sum[maxn];
int main()
{
//freopen("D:\\common_text\\code_stream\\in.txt","r",stdin);
//freopen("D:\\common_text\code_stream\\out.txt","w",stdout);
int q;
int n,k;
gbtb;
cin>>q;
while(q--)
{
cin>>n>>k;
cin>>s;
string str1,str2,str3;
str1="";
str2="";
str3="";
for(int i=0;i<n;++i)
{
if(i%3==0)
{
str1.pb('R');
}else if((i%3)==1)
{
str1.pb('G');
}
else
{
str1.pb('B');
}
}
for(int i=0;i<n;++i)
{
if(i%3==0)
{
str2.pb('G');
}else if((i%3)==1)
{
str2.pb('B');
}
else
{
str2.pb('R');
}
}
for(int i=0;i<n;++i)
{
if(i%3==0)
{
str3.pb('B');
}else if((i%3)==1)
{
str3.pb('R');
}
else
{
str3.pb('G');
}
}
int ans=inf;
rep(i,0,n)
{
if(s[i]!=str1[i])
{
a[i]=1;
}else
{
a[i]=0;
}
}
sum[0]=a[0];
rep(i,1,n)
{
sum[i]=sum[i-1]+a[i];
}
for(int i=k-1;i<n;++i)
{
int temp=0;
if(i-k>=0)
temp=sum[i-k];
ans=min(ans,sum[i]-temp);
}
rep(i,0,n)
{
if(s[i]!=str2[i])
{
a[i]=1;
}else
{
a[i]=0;
}
}
sum[0]=a[0];
rep(i,1,n)
{
sum[i]=sum[i-1]+a[i];
}
for(int i=k-1;i<n;++i)
{
int temp=0;
if(i-k>=0)
temp=sum[i-k];
ans=min(ans,sum[i]-temp);
}
rep(i,0,n)
{
if(s[i]!=str3[i])
{
a[i]=1;
}else
{
a[i]=0;
}
}
sum[0]=a[0];
rep(i,1,n)
{
sum[i]=sum[i-1]+a[i];
}
for(int i=k-1;i<n;++i)
{
int temp=0;
if(i-k>=0)
temp=sum[i-k];
ans=min(ans,sum[i]-temp);
}
cout<<ans<<endl;
}
return 0;
}
inline void getInt(int* p) {
char ch;
do {
ch = getchar();
} while (ch == ' ' || ch == '\n');
if (ch == '-') {
*p = -(getchar() - '0');
while ((ch = getchar()) >= '0' && ch <= '9') {
*p = *p * 10 - ch + '0';
}
}
else {
*p = ch - '0';
while ((ch = getchar()) >= '0' && ch <= '9') {
*p = *p * 10 + ch - '0';
}
}
}
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