Language: C++

/*                                  /   \                                                                   (Hello)//JSM//*/      
/*                                 /*****\                                                                              */
/*     jai shree mataji           /       \ |\| |_| $ |-| K                                                            */
#include<vector>
#include<iostream>
#include<algorithm>                
#include<string.h>                 
#include<bits/stdc++.h>            
#include<cstring>                  
#include<map>                     
#include<cmath>                  
#include<set>
using namespace std;            
#define ll long long int        
#define ld long double
#define MOD 1000000007 
#define PI 3.1415926535897932384626433832795                                               
#define YY cout<<"YES"
#define NN cout<<"NO"
#define EE cout<<"\n"
#define ee cout<<"\n"
#define ne cout<<"-1"
#define pb push_back 
#define ff first
#define ss second
#define mkp make_pair    
#define mkt make_tuple
#define all(v ) sort(v.begin(), v.end());      
#define resrt(v ) sort(v.rbegin(), v.rend()); 
#define fo(i , n) for(ll i = 0 ; i < n ; i++)
#define vi vector<ll> 
#define pr pair<ll,ll>
#define tpl tuple<ll,ll,ll>
#define mpl map<ll,ll>
#define mpc map<char,ll>
#define mps map<string,ll>
bool sortbysec(const pair<int,int> &a,const pair<int,int> &b) {return (a.second < b.second); }
ll containsDigit(ll number, ll digit){while (number != 0){int curr_digit = number % 10;if (curr_digit == digit) return 1;number /= 10;}return 0;}
ll soltn(ll x, ll n) {if(x>=0) return x%n;    else return n+x%n;}
ll bin_srch(ll L[], ll N,ll item){ll flag=-2;int l=0,u=N-1,m;while(l<=u) {m=(l+u)/2;if(item==L[m]) {flag=m-1; break;} else if(item>L[m]) l=m-1; else u=m+1;} return flag;}
ll pdiv(ll x,ll y){if(x%y==0) return x/y;else return x/y+1;}
ll coda(ll n,ll d){ll count=0;while(n!=0){if(n%10==d)count++;n/=10;}return count;}
ll minDissame(ll a[], ll n)  {map<ll,ll> hmap; ll minDistance = INT_MAX;  ll previousIndex = 0, currentIndex = 0;  for (ll i = 0; i < n; i++) {  if (hmap.find(a[i])!=hmap.end()) {  currentIndex = i;     previousIndex = hmap[a[i]];   minDistance = min((currentIndex -        previousIndex),minDistance);        }hmap[a[i]] = i;  }    return (minDistance == INT_MAX ? -1 : minDistance);  } 
ll minDistanytwo(ll arr[], ll n, ll x, ll y) {ll p = -1, min_dist = INT_MAX; for(ll i=0 ; i<n ; i++) {if(arr[i]==x || arr[i]==y) {if( p != -1 && arr[i] != arr[p])  min_dist = min(min_dist , i-p); p=i;  }} if(min_dist==INT_MAX) return -1; return min_dist; } 
ll mdls(ll x){return max(x,-x);}
ll pow1(ll n,ll p){if(p==0)return 1;ll x=pow1(n, p/2);x=(x*x)%MOD;if(p%2==0) return x;else return (x*n)%MOD;}
string fumn(string s){string p;for(ll i = 0; i < s.length(); i++)if(isalpha(s[i]))p+=toupper(s[i]);return p;}
ll turns(string S,string SS){ ll i,cc=0,l=S.length();fo(i,l){ if(S[i]!=SS[i]) cc++; } return cc/2;}
ld polygonArea(ld X[], ld Y[], ll n) {  double area = 0.0; ll j = n - 1; for (ll i = 0; i < n; i++) {   area += (X[j] + X[i]) * (Y[j] - Y[i]);  j = i;  }  return abs(area / 2.0); } 
ll isSubstr(string s1, string s2) {   ll M = s1.length();    ll N = s2.length();  for (ll i = 0; i <= N - M; i++) { ll j; for (j = 0; j < M; j++) if (s2[i + j] != s1[j])  break;         if (j == M) return i;   } return -1; } 
//s2>s1;if(not) rtrn -1;
ll hhh2(unsigned ll n) {   if (n < 1)  return 0; ll res = 1;  ll u=0; for (ll i=0; i<8*sizeof(unsigned ll); i++) {    ll curr = 1 << i; if (curr > n)      break;   u++;  res = curr;} return u; }
ld ldgcd(ld a,ld b){ return a<0.0001?b:ldgcd(fmod(b,a),a);} 
void onesol(ll a,ll b,ll& x,ll& y,ll& g) {if(!b) {g=a;x=1;y=0;} else{onesol(b,a%b,y,x,g);y-=a/b*x;}}
/*struct pp
{
	ll x,y;
};*/
/*
https://www.geeksforgeeks.org/prime-factorization-using-sieve-olog-n-multiple-queries/*/
/*PLEASE START @ANU$HK GAUTAM
KEEP PATIENCE AND CONFIDENCE*/
int main(){
ios_base::sync_with_stdio(0);
cin.tie(0);		  ll t=1;
      
	   cin>>t;
      
	      while(t--)
{
/*
6
6 1
100010
6 2
000000
5 1
10101
3 1
001
2 2
00
1 1
0
*/
/*
/*
6
6 1
100010
0 1 5 7
1 4 2  
6 2
000000
5 1
10101
3 1
001
2 2
00
1 1
0
*/
ll n,k,i,j,l;
cin>>l>>k;
string s;
cin>>s;
ll cc=0,f=0,ans=0;
vi v;
  //12   3
// 1000 0 00010010100010
for(i=0;i<l;i++)
{
	if(s[i]=='0')
	{
		cc++;
		j=1;
	}
	else
	{
		f++;
		if(f==2&&cc>2*k)
		{
			ans+=(cc-2*k-1)/(k+1)+1;
		}
		else if(f==1&&cc>k)
		{
			ans+=(cc-k-1)/(k+1)+1;
		} 
		cc=0;
		f=1;
		j=-1;
	}
}
if(j==1&&cc>k)
		{
			ans+=(cc-1)/(k+1)+1;
		}
		//9 4
		//000000000
if(f==0)
ans=(l-1)/(k+1)+1;
cout<<ans;
ee;
	
}
return 0;
}
�


Compiler output:
a.cc:149:1: error: '\U0000fffd' does not name a type
  149 | �
      | ^