 # SVRT - Editorial

Author: Akash Bhalotia
Tester: Rahul Dugar
Editorialist: Aman Dwivedi

Easy

# PREREQUISITES:

Observation, Maths

# PROBLEM:

You wish to place K servers across N locations. The locations are arranged in a circular manner. You want to place the servers such that the maximum clockwise distance between any two adjacent servers is as less as possible. Find this minimum possible distance that can be achieved, and also the minimum number of pairs of servers separated by this distance.

# EXPLANATION:

Let’s visualize this problem as we have a circle of length N and we made K cuts on this circle. Then we divide the circle from where the cuts had been made. Hence we will get K pieces of the circle. We need to minimize the maximum length possible of any piece.

The optimal way to divide the circle into K pieces is to divide it into the pieces of length N/K. But after dividing it into K pieces there might be some residual that is still left. Hence we will have two cases:

CASE 1 : When (N mod K=0)

• In this case when N is completely divisible by K, then there will be no residual left after dividing the circle into K pieces of length N/K. In this case the maximum length that is possible is N/K and the number of pieces of this length are K.

CASE 2: When (N mod K \neq 0)

• In this case, when N is not completely divisible by K then after dividing the circle into K pieces of length N/K, there will be a residual left of length (N mod K).

• Since K is always greater than (N mod K). Hence from the residual, we will give unit length to (N mod K) pieces such that the length of (N mod K) pieces will increase by a unit length.

• Hence, In this case the maximum length that is possible is (N/K+1) and the number of pieces of this length are (N mod K).

# TIME COMPLEXITY:

O(1) per test case

# SOLUTIONS

Setter
``````//created by Whiplash99
import java.io.*;
import java.util.*;
class A
{
public static void main(String[] args) throws Exception
{

int i,N;

StringBuilder sb=new StringBuilder();

while (T-->0)
{
N=Integer.parseInt(s);
int K=Integer.parseInt(s);

int D=(N+K-1)/K;
int X=(N%K==0)?K:(N-K*(N/K));

sb.append(D).append(" ").append(X).append("\n");
}
System.out.println(sb);
}
}
``````
Tester
``````#pragma GCC optimize("Ofast")
#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 double f80;
typedef pair<int, int> pii;
typedef pair<double, double> pdd;
typedef pair<ll,ll> pll;
#define double long double
#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 int infi=0x3f3f3f3f;
const ll infl=0x3f3f3f3f3f3f3f3fLL;
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>0) {
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 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){
}
}
}

void solve() {
int lol=(n-1)/k+1;
cout<<lol<<" "<<(n-1)%k+1<<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(1);
//	cin>>t;
fr(i,1,t)
solve();
assert(getchar()==EOF);
#ifdef rd
cerr<<endl<<endl<<endl<<"Time Elapsed: "<<((double)(clock()-clk))/CLOCKS_PER_SEC<<endl;
#endif
}

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

#define int long long

void solve()
{
int n,k;
cin>>n>>k;

int d=n/k;
int x=k;

if(n%k!=0)
{
d++;
x=n%k;
}

cout<<d<<" "<<x<<"\n";
}

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

int t;
cin>>t;

while(t--)
solve();

return 0;
}

``````
4 Likes

Can anyone tell me how to not exceed time limit in java?

pro tip
write efficient code!!! xd

https://www.codechef.com/viewsolution/64297796
Help!!! can someone tell me what is wrong in my code thanks in advance

hey @manav2711