# SMOL - Editorial

Author: Ashish Gupta
Tester: Rahul Dugar
Editorialist: Aman Dwivedi

Cakewalk

Maths

# PROBLEM:

You are given two integers N and K. Your task is to find the smallest whole number value of N that can be obtained by operating N=N−K any number of times.

# QUICK EXPLANATION:

The smallest whole number that can be obtained is the remainder when N is divided by K.

# EXPLANATION:

We need to find the smallest whole number that can be obtained by operation N=N-K any number of times. Since we can operate any number of times, we say that our answer will be N-(X*K), such that:

N-(X*K) \ge 0
N \ge X*K
X \le N/K

Since our goal is to minimize the value of the whole number, that means we need to maximize the value of X. The maximum value that X can take is \lfloor \frac{N}{K} \rfloor. That is nothing but the quotient when N is divided by K. Hence our answer will be:

N- \lfloor \frac{N}{K} \rfloor * K

which is nothing but just (N % K). Hence the smallest whole number that can be obtained is the remainder when N is divided by K.

# TIME COMPLEXITY:

O(1) per test case.

# SOLUTIONS:

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

#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define endl "\n"
#define int long long

const int N = 3005;

int32_t main()
{
IOS;
int t;
cin >> t;
while(t--)
{
int n, k;
cin >> n >> k;
if(k == 0)
cout << n << endl;
else
cout << n % k << endl;
}
return 0;
}

Tester
#include <bits/stdc++.h>
using namespace std;
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/rope>
using namespace __gnu_pbds;
using namespace __gnu_cxx;
#ifndef rd
#define trace(...)
#define endl '\n'
#endif
#define pb push_back
#define fi first
#define se second
#define int long long
typedef long long ll;
typedef long double f80;
#define double long double
#define pii pair<int,int>
#define pll pair<ll,ll>
#define sz(x) ((long long)x.size())
#define fr(a,b,c) for(int a=b; a<=c; a++)
#define rep(a,b,c) for(int a=b; a<c; a++)
#define trav(a,x) for(auto &a:x)
#define all(con) con.begin(),con.end()
const ll infl=0x3f3f3f3f3f3f3f3fLL;
const int infi=0x3f3f3f3f;
//const int mod=998244353;
const int mod=1000000007;
typedef vector<int> vi;
typedef vector<ll> vl;

typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> oset;
auto clk=clock();
mt19937_64 rang(chrono::high_resolution_clock::now().time_since_epoch().count());
int rng(int lim) {
uniform_int_distribution<int> uid(0,lim-1);
return uid(rang);
}
int powm(int a, int b) {
int res=1;
while(b) {
if(b&1)
res=(res*a)%mod;
a=(a*a)%mod;
b>>=1;
}
return res;
}

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 {
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) {
}

void solve() {
if(k==0)
cout<<n<<endl;
else
cout<<n%k<<endl;
}
signed main() {
ios_base::sync_with_stdio(0),cin.tie(0);
srand(chrono::high_resolution_clock::now().time_since_epoch().count());
cout<<fixed<<setprecision(10);
//	int t;
//	cin>>t;
fr(i,1,t)
solve();
#ifdef rd
cerr<<endl<<endl<<endl<<"Time Elapsed: "<<((double)(clock()-clk))/CLOCKS_PER_SEC<<endl;
#endif
}

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

int32_t main()
{
ios_base::sync_with_stdio(0);
cin.tie(0);

int t;
cin>>t;

while(t--)
{
int n,k;
cin>>n>>k;

if(k==0)
{
cout<<n<<"\n";
continue;
}

cout<<(n%k)<<"\n";
}

return 0;
}


1 Like