CF1353E K-periodic Garland（贪心/dp）

Question

有n盏灯，0代表暗，1代表亮，相邻两个1之间为周期k，求出最少的改变次数

Solution

First

贪心方法

详见博客https://blog.csdn.net/cheng__yu_/article/details/106145027

#include <iostream>
#include <cstdio>
#include <fstream>
#include <algorithm>
#include <cmath>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <cstring>
#include <map>
#include <stack>
#include <set>
#define maxn 1005
#define inf 0x3f3f3f
#define endl '\n'
#pragma GCC optimize(2)
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
typedef long long ll;
//ll fpm(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
//ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}
//ll fastPow(ll a,ll b) {ll res=1; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a;a=a*a;}return res;}
int t;
int n,k;
string s;
int main(){
ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
cin>>t;
while(t--){
	cin>>n>>k;
	cin>>s;
	int cnt=count(s.begin(),s.end(),'1');
	int anss=inf;
	for(int i=0;i<k;i++){
		int temp=0;
		for(int j=i;j<n;j+=k){
			if(s[j]=='0') temp++;
			else temp--;
			temp=min(temp,0);
			anss=min(anss,cnt+temp);
		}
	}
	cout<<anss<<endl;
}
	return 0;
}

Second

#include <iostream>
#include <cstdio>
#include <fstream>
#include <algorithm>
#include <cmath>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <cstring>
#include <map>
#include <stack>
#include <set>
#define maxn 1005
#define inf 0x3f3f3f
#define endl '\n'
#pragma GCC optimize(2)
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
typedef long long ll;
//ll fpm(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
//ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}
//ll fastPow(ll a,ll b) {ll res=1; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a;a=a*a;}return res;}
int t;
int n,k;
string s;
int dp[1000050][2];
int pre[1000050];
int main(){
ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
cin>>t;
while(t--){
	cin>>n>>k;
	cin>>s;
	s="0"+s;
	for(int i=1;i<=n;i++){
		pre[i]=pre[i-1]+(s[i]=='1');
	}
	for(int i=1;i<=n;i++){
		int z=max(0,i-k);
		dp[i][0]=min(dp[i-1][1],dp[i-1][0])+(s[i]=='1');
		dp[i][1]=min(dp[z][1]+pre[i-1]-pre[z],pre[i-1])+(s[i]=='0');
	}
	int ans=min(dp[n][0],dp[n][1]);
	cout<<ans<<endl;
}
	return 0;
}