# EXPIRY-Editorial

Setter: Hriday
Tester: Nishank Suresh, Satyam
Editorialist: Devendra Singh

Cakewalk

None

# PROBLEM:

Eikooc loves bread. She has N loaves of bread, all of which expire after exactly M days. She can eat upto K loaves of bread in a day. Can she eat all the loaves of bread before they expire?

# EXPLANATION:

All the loaves of bread expire exactly after M days. Therefore the order of eating loaves is not important here. The best startegy would be to maximise the number of loaves she can eat before all of them expire. She has M days and can eat upto K loaves a day. So, the maximum number of loaves she can eat is M\cdot K. If this value is more than or equal to N the answer is Yes otherwise No.

# TIME COMPLEXITY:

O(1) for each test case.

# SOLUTION:

Tester-1's Solution(Python)
for _ in range(int(input())):
N, M, K = map(int, input().split())
print('YES' if K*M >= N else 'NO')

Tester-2's Solution
#include <bits/stdc++.h>
using namespace std;
#ifndef ONLINE_JUDGE
#define debug(x) cerr<<#x<<" "; _print(x); cerr<<nline;
#else
#define debug(x);
#endif

/*
------------------------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){
}
long long readIntLn(long long l,long long r){
}
string readStringLn(int l,int r){
}
string readStringSp(int l,int r){
}

/*
------------------------Main code starts here----------------------------------
*/
int MAX=100000;
void solve(){
if(m*k>=n){
cout<<"Yes\n";
}
else{
cout<<"No\n";
}
return;
}
int main(){
ios_base::sync_with_stdio(false);
cin.tie(NULL);
#ifndef ONLINE_JUDGE
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
freopen("error.txt", "w", stderr);
#endif
while(test_cases--){
solve();
}
assert(getchar()==-1);
return 0;
}

Editorialist's Solution
#include "bits/stdc++.h"
using namespace std;
#define ll long long
#define pb push_back
#define all(_obj) _obj.begin(),_obj.end()
#define F first
#define S second
#define pll pair<ll, ll>
#define vll vector<ll>
const int N=1e5+11,mod=1e9+7;
ll max(ll a,ll b) {return ((a>b)?a:b);}
ll min(ll a,ll b) {return ((a>b)?b:a);}
void sol(void)
{
int n,m,k;
cin>>n>>m>>k;
if(n>m*k)
cout<<"NO\n";
else
cout<<"YES\n";
return ;
}
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL),cout.tie(NULL);
int test=1;
cin>>test;
while(test--) sol();
}