# Chef and demonetization problem

I don’t know why i got WA, it passes the possible test cases as much as i know
https://ide.geeksforgeeks.org/fDTvxwBHtc

1
28 10
answer should be 3 but u are giving 6
you can use any even number in range 0…n

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``````#include <iostream>
#define ll long long
using namespace std;
int main(){
int t; cin >> t;
while(t--){
ll s; cin >> s;
ll n; cin >> n;
ll l2=0,total=0;
if(s%2==0){
int l = s/n;
int total1 = s-(l*n);
if(total1!=0) cout << l+1 << "\n";
else cout << l << "\n";
}else{
total = s/n;
l2 = s-(total*n);
if(l2!=1) total+=l2/(l2-1);
else l2=2;
std::cout << total+1 << "\n";
}
}
}
``````

My code might help

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But why 3 ?

thanks

you can use coins of 10 10 8

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You could use two 10 and one 8

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confused!

Well if you are trying to minimize the coins used you have to use the maximum number of even valued coins possible everytime and if it is odd just add 1 coz we can use it.

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#include
using namespace std;

int main(){
int t;
cin>>t;
while(t–){
int s, n;
cin>>s>>n;
int ans = 0;
if(n==1){
cout<<s<<"\n";
continue;
}
while(s>1){
ans += (s/n);
s %= n;
n -= 2;
}
if(s>0){
ans++;
}
cout<<ans<<"\n";
}
return 0;
}

I don’t know why my code is giving TLE. It is passing all potential test cases.

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The worst case number is 10^9 so you can’t use brute force .

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