Hello coders, This is my first post on CodeChef discuss and the post contains solutions to our contest problems which we conducted in our college named Coding Maniacs on 14/05/2020. So if you found any mistakes do suggest me.
Contest Page: Contest Page | CodeChef
1. Two Number Game
Solution :
//coded by vishal mourya
#include<bits/stdc++.h>
#define fasthoja ios_base::sync_with_stdio(false); cin.tie(NULL);
using namespace std;
typedef long long int ll;
int main(void){
fasthoja;
ll t; cin>>t;
while(t–){
ll a,b;
cin>>a>>b;
if( a == b ) cout<<"YES\n";
else if( a > b ) cout<<"NO\n";
else cout<<"YES\n";
}
return 0;
}
2. Helping Machine
Solution :
//coded by vishal mourya
#include<bits/stdc++.h>
using namespace std;
typedef long long int ll;
int main(void){
ll t; cin>>t;
while(t–){
ll n; cin>>n;
vector v(n);
ll t1=0,t2=0,t3=0;
for(ll i = 0 ; i < n ; i++ ){
cin>>v[i];
if(v[i] == 1 || v[i] == 2)
t1++;
else if(v[i] == 3 || v[i] == 4)
t2++;
else if(v[i] == 5 || v[i] == 6)
t3++;
}
cout<<"TYPE 1 - "<<t1<<"\n";
cout<<"TYPE 2 - "<<t2<<"\n";
cout<<"TYPE 3 - "<<t3<<"\n";
}
return 0;
}
3. Rail Yatra
Solution :
#include <bits/stdc++.h>
using namespace std;
typedef double ld;
vector x, v;
int main()
{
cout.precision(31);
cin.tie(0);
ios_base::sync_with_stdio(false);
int n;
cin >> n;
x.resize(n);
v.resize(n);
for (int i = 0; i < n; i++)
cin >> x[i];
for (int i = 0; i < n; i++)
cin >> v[i];
ld l = 0, r = 1e10;
for (int ui = 0; ui < 100; ui++)
{
ld med = (r + l) / (ld)(2);
ld left = -1e12, right = 1e10;
for (int i = 0; i < n; i++)
{
left = max(left, x[i] - v[i] * med);
right = min(right, x[i] + v[i] * med);
}
if (right >= left)
r = med;
else
l = med;
}
cout << fixed<<setprecision(30)<<l<<endl;
return 0;
}
4. Chef Birthday
Solution :
/*Author Sanjay shukla */
#include <bits/stdc++.h>
using namespace std;
int main()
{
int testcase;
int count,min,max;
cin>>testcase;
while((testcase–)!=0)
{
int n,len=0,i,p;
cin>>n>>p;
string a[n],query[p];
//entering a sequence of relatives
for(i=0;i<n;i++)
cin>>a[i];
for(i=0;i<p;i++)
cin>>query[i];
//concatenation of strings
int j;
for(j=0;j<p;j++)
{
max=0,min=1000000;
for(i=0;i<n;i++)
{ count=0;
string result=a[i]+query[j];
//cout<<result<<endl;
int l=result.length();//length of concatenated string
len=l%10;
//cout<<len<<" ";
// this while block checks how many steps is required to convert concatenated into the first composite number(ie. FOUR)
while(len!=4)
{
switch(len)
{
case 0:
len=4;
count++;
break;
case 1:
len=3;
count++;
break;
case 2:
len=3;
count++;
break;
case 3:
len=5;
count++;
break;
case 4:
len=4;
count++;
break;
case 5:
len=4;
count++;
break;
case 6:
len=3;
count++;
break;
case 7:
len=5;
count++;
break;
case 8:
len=5;
count++;
break;
case 9:
len=4;
count++;
break;
}//end of switch
}//end of inner while loop
//calculating minimum and maximum of count
if(count<min)
{
min=count;
}
if(count>max)
max=count;
cout<<count<<" ";//this cout display all the count values for every single query on a single line
}//end of array
cout<<endl;//this endl seperates count values for every single quer
cout<<min<<" "<<max<<endl;
}//end of while loop of query
}//end of testcase loop
}
5. Ticket Wala
Solution :
/*
coded by addy
*/
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef vector vec;
#define fl(i,n) for(i=0;i<n;i++)
#define rfl(i,n) for(i=n-1;i>=0;i–)
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
ll n,i,sum1,sum2,t,even,odd,m1,m2,k,l,min1,min2,v,avg1,avg2;
cin>>t;
while(t–)
{
cin>>n;
vec a(n),b(n),c(10),f(10),s(10);
sum1=0;sum2=0;
fl(i,n)
{
cin>>a[i];//taking input of all the unit places of the tickets
}
fl(i,n)
{
cin>>b[i];//taking input of price of each ticket respectively
}
fl(i,10)
{
cin>>c[i];//tax on type of each digit
}
fl(i,n)
{
f[a[i]]++;//frequency of each digit
s[a[i]]+=b[i];//total money spent on ticket of each digit type
}
even=0;odd=0;m1=0;m2=0;
fl(i,10)
{
if((i%2)==0)
{
m1+=c[i];//calculation of sum for mean of tax of even numbers
even+=f[i];
}
else
{
m2+=c[i];//calculation of sum for mean of tax of odd numbers
odd+=f[i];
}
}
min1=n;min2=n;k=0;l=0;
if(even!=0&&odd!=0)
{
fl(i,10)
{
if(i%2==0)
{
if((f[i]<min1)&&(f[i]>0))
{
min1=f[i];
k=i;
}
}
else
{
if((f[i]<min2)&&(f[i]>0))
{
min2=f[i];
l=i;
}
}
}
avg1=(s[k]/f[k]);//avg price of no. with minimum frequency in even digit
avg2=(s[l]/f[l]);//avg price of no. with minimum frequency in odd digit
}
m1/=5;m2/=5;
fl(i,10)
{
s[i]=s[i]-f[i]*c[i];//calculating (sum of amount of each digit)-(tax on each ticket of each digit)
if(i%2!=0)
sum1+=s[i];//calculated by chef (odd units digits amount sum)
else
sum2+=s[i];//calculated by cheffa (even unit digits amount sum)
}
if(even==odd)
{
cout<<“0”<<endl;
cout<<sum1<<" “<<sum2<<endl;//output
}
else if(even==0||odd==0)
{
cout<<“0”<<endl;
cout<<sum1<<” “<<sum2<<endl;//output
}
else
{
if(even>odd)
v=(even-odd)*avg2-(even-odd)*m2;//virtual sum
else
v=(odd-even)*avg1-(odd-even)*m1;//virtual sum
cout<<“1”<<endl;
cout<<sum1+sum2<<” "<<v+sum1+sum2<<endl;
}
}
return 0;
}
Hoping to conduct more such contest in our college 
Thank You CodeChef for creating such a wonderful environment.