# CYBV - Editorial

Practice

Author: Jenish Monpara
Tester: Smit Mandavia

CAKEWALK

# PREREQUISITES:

No prerequisites, simple math

# PROBLEM:

We have to distribute K weapons among N kid cyborgs such that the difference between kid cyborg having the maximum weapons and the kid cyborg having minimum weapons should be less than or equal to 1 after distribution of all the weapons. We want to know the minimum number of weapons.

# QUICK EXPLANATION:

We can give \frac{K}{N} weapons to each of the kid cyborgs. The remaining K \% N weapons can be given one each to K \% N kids. (\% is the modulo operation)

# EXPLANATION:

Firstly, we can notice that we always can distribute K- K \% N (where \% is the modulo operation) weapons evenly between kids, with each kid getting \frac{K}{N} weapons. Now the remaining weapons are K \% N, which can be given 1 to K \% N kids each. Thus, if K\%N is non-zero, the difference between kids with maximum weapons and kid with minimum weapons would be one. Otherwise, it would be 0.

# SOLUTIONS:

Setter's Solution
#include <bits/stdc++.h>
#define int long long
#define fio ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0)

using namespace std;

int32_t main()
{
fio;
int t;
cin>>t;
while (t--)
{
int n,k;
cin>>n>>k;
cout<<k/n<<"\n";
}
return 0;
}

Tester's Solution

#include<bits/stdc++.h>
using namespace std;

#define ll long long int
#define FIO ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define mod 1000000007

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;
}
assert(l<=x && x<=r);
return x;
} else {
cerr << (int)g << "\n";
assert(false);
}
}


}
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){
}
long long readIntLn(long long l,long long r){
}
}
}

int main()
{
ll t,n,q,k,i,j,sum_n,sum_q;
cin >> t;
while(t–){
cin >> n >> k;
cout << k/n << “\n”;
}
return 0;
}

Editorialist's Solution
#include <bits/stdc++.h>
#define ll long long

using namespace std;

int main()
{
ll t,n,k;
cin>>t;
for(int i = 0; i < t; i++)
{
cin>>n>>k;
cout<<k/n<<endl;
}
return 0;
}

1 Like

Can someone please explain why this solution is correct?

I realized that for input, say n=10 and k=10, answer should be 1 but in my case it was 10 and still solution worked. Am I missing something or there is lack of test cases?

Probably there are no testcases such that n==k
If n is not equal to k the answer will be correct

#include <bits/stdc++.h>

using namespace std;

#define ll long long

int main() {
int t;
cin>>t;
while (t–) {
ll n,k;
cin>>n>>k;
if (n==k) {
printf("%lld\n",n);
continue;
}
if (n>k) {
printf(“0\n”);
continue;
}
printf("%lld\n",k/n);
}
return 0;
}

*** bro here you have written n==k
consider n=5 k=5
then one kid will have only one weapon that is minimum weapon=1
so you have to remove this condition

If n=7 and k=19
we can distribute as 2 2 2 2 2 5 4
In this case answer will be 1
Here it is not mentioned we can use only two types of numbers.
Can anyone clear me out here?

In the case you have considered the difference between max and min weapons is 5-2=3 not 1 or 0 as mentioned in the question…The distribution would be like 3 3 3 3 3 2 2