PROBLEM LINK:
Practice
Contest: Division 1
Contest: Division 2
Contest: Division 3
Authors: Daanish Mahajan
Testers: Abhinav Sharma and Lavish Gupta
Editorialist: Nishank Suresh
DIFFICULTY:
Cakewalk
PREREQUISITES:
None
PROBLEM:
Chef has a work plan for the next N hours, and for each hour he knows if he must work or can take a break. He can take a nap if there are at least K consecutive hours of break time. What is the maximum number of naps he can take?
EXPLANATION:
Suppose there are x hours of consecutive break time. Chef needs at least K hours to take a nap, so in x hours, he can take at most \lfloor \frac{x}{K} \rfloor naps, where \lfloor a \rfloor denotes the floor of a, i.e, the largest integer less than or equal to a.
So, the maximum number of naps Chef can take can be computed as follows:
- Let the (maximal) segments of break time in Chef’s schedule have lengths x_1, x_2, \ldots, x_m (a segment is said to be maximal if it is not contained in another segment of break time which has strictly larger length).
- The maximum number of naps is then \lfloor \frac{x_1}{K} \rfloor + \lfloor \frac{x_2}{K} \rfloor + \ldots + \lfloor \frac{x_m}{K} \rfloor
For example, if Chef’s schedule is '0010110001110000' and K = 2, the maximal segments of free time have lengths 2, 1, 3, 4 from left to right. The maximum number of naps is then \lfloor \frac{2}{2} \rfloor + \lfloor \frac{1}{2} \rfloor + \lfloor \frac{3}{2} \rfloor + \lfloor \frac{4}{2} \rfloor = 1 + 0 + 1 + 2 = 4.
In most languages, computing \lfloor \frac{x}{K} \rfloor can be done by integer division, that is, just x / K. Note that python users need to use x // K instead.
Finding the maximal segments of break time can be done in \mathcal{O}(|S|) by iterating over the string and keeping a variable denoting the length of the current segment, as in the following pseudocode:
cur = 0
for i from 1 to N
if S[i] = '0'
cur += 1
else
// cur is now the length of a maximal segment, do whatever you want with it
// ...
// reset the length to 0
cur = 0
end
// at the end of the loop, cur holds the length of the maximal segment of break time which is a suffix of S
TIME COMPLEXITY:
\mathcal{O}(N) per test case.
SOLUTIONS:
Tester's Solution (C++)
#include <bits/stdc++.h>
using namespace std;
/*
------------------------Input Checker----------------------------------
*/
long long readInt(long long l,long long r,char endd){
long long x=0;
int cnt=0;
int fi=-1;
bool is_neg=false;
while(true){
char g=getchar();
if(g=='-'){
assert(fi==-1);
is_neg=true;
continue;
}
if('0'<=g && g<='9'){
x*=10;
x+=g-'0';
if(cnt==0){
fi=g-'0';
}
cnt++;
assert(fi!=0 || cnt==1);
assert(fi!=0 || is_neg==false);
assert(!(cnt>19 || ( cnt==19 && fi>1) ));
} else if(g==endd){
if(is_neg){
x= -x;
}
if(!(l <= x && x <= r))
{
cerr << l << ' ' << r << ' ' << x << '\n';
assert(1 == 0);
}
return x;
} else {
assert(false);
}
}
}
string readString(int l,int r,char endd){
string ret="";
int cnt=0;
while(true){
char g=getchar();
assert(g!=-1);
if(g==endd){
break;
}
cnt++;
ret+=g;
}
assert(l<=cnt && cnt<=r);
return ret;
}
long long readIntSp(long long l,long long r){
return readInt(l,r,' ');
}
long long readIntLn(long long l,long long r){
return readInt(l,r,'\n');
}
string readStringLn(int l,int r){
return readString(l,r,'\n');
}
string readStringSp(int l,int r){
return readString(l,r,' ');
}
/*
------------------------Main code starts here----------------------------------
*/
const int MAX_T = 1e5;
const int MAX_N = 1e5;
const int MAX_SUM_LEN = 1e5;
#define fast ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define ff first
#define ss second
#define mp make_pair
#define ll long long
int sum_len = 0;
int max_n = 0;
int yess = 0;
int nos = 0;
int total_ops = 0;
const ll MX=200000;
ll fac[MX], ifac[MX];
const ll mod = 998244353;
ll po(ll x, ll n ){
ll ans=1;
while(n>0){
if(n&1) ans=(ans*x)%mod;
x=(x*x)%mod;
n/=2;
}
return ans;
}
void solve()
{
ll n,x;
n = readIntSp(1,MAX_N);
x = readIntLn(1,n);
string s;
s = readStringLn(1,MAX_N);
int cur = 0, ans = 0;
for(auto h:s){
assert(h=='0' || h=='1');
if(h=='0') cur++;
else cur = 0;
if(cur == x){
ans++;
cur = 0;
}
}
cout<<ans<<'\n';
}
signed main()
{
#ifndef ONLINE_JUDGE
freopen("input.txt", "r" , stdin);
freopen("output.txt", "w" , stdout);
#endif
fast;
int t = 1;
t = readIntLn(1,MAX_T);
for(int i=1;i<=t;i++)
{
solve();
}
assert(getchar() == -1);
cerr<<"SUCCESS\n";
cerr<<"Tests : " << t << '\n';
cerr<<"Sum of lengths : " << sum_len << '\n';
// cerr<<"Maximum length : " << max_n << '\n';
// cerr<<"Total operations : " << total_ops << '\n';
//cerr<<"Answered yes : " << yess << '\n';
//cerr<<"Answered no : " << nos << '\n';
}
Editorialist's Solution (Python)
for _ in range(int(input())):
n, k = map(int, input().split())
s = input() + '1'
cur, ans = 0, 0
for c in s:
if c == '1':
ans += cur//k
cur = 0
else:
cur += 1
print(ans)