Not possible. I think.
anyone who solved problem D? , I was getting a slight difference in calculated revenue also in the problem when number of patients are greater than total number of rooms was not defined.
@shivam_js
For problem D take into account the following observations:
For the year:-
Before august-
odd months have 31 days and even months have 30 days except February it has 28 days(non-leap year).
August onwards-
odd months have 30 days and even months have 31 days.
For calculating revenue for each day:-
There are 3 conditions
if(no. of patients > = no. of rooms) //all rooms will be accquired
revenue+=(no. of tv rooms * cost of tv rooms) + (no. of non-tv rooms * cost of non-tv rooms)
else if(no. of patients < no. of rooms && no. of patients <= no. of non-tv rooms)
//in this the patients will acquire only non-tv rooms
revenue+=(no. of patients * cost of non-tv rooms)
else if(no. of patients < no of rooms && no. of patients > no. of non-tv rooms)
//in this all non-tv rooms will be acquired and remaining patients will acquire tv rooms
//remaining patients=no. of patients - no. of non-tv rooms
revenue+=(no. of non-tv rooms * cost of non-tv rooms) + (remaining patients * cost of tv rooms)
@akashdileep Can you please review this function i made to calculate the revenue
def check(tvrooms,totalrooms,r1,r2):
revenue=0
month=[31,28,31,30,31,30,31,31,30,31,30,31]
nontvrooms=totalrooms-tvrooms
for i in range(1,13):
for j in range(1,month[i-1]+1):
people=((6-i)**2)+abs(j-15)
print(tvrooms,nontvrooms)
if people>=totalrooms:
revenue+=(tvrooms*r1)+(nontvrooms*r2)
else:
if people<=nontvrooms:
revenue+=nontvrooms*r2
else:
revenue+=((people-nontvrooms)*r1)+(nontvrooms*r2)
return revenuecases = int(input())
for i in range(1, cases + 1):
value = int(input())
coincount = 0
while value >= 1:
value = value // 2
coincount = coincount + 1
print(coincount)
i am sorry
So, there are no fault with the test cases
Lol only if you looked at the SS above, which says n2-n1 >= 35 clearly.
Same bro!
yah i reconsidered it. there were actually lot of problem with the test cases of tcs actually
1 Like
like the test cases of 3rd problem was also unprecedented
else:
if people<=nontvrooms:
revenue+=people*r2
//Because no. of people can be less than no. of non-tv rooms
//revenue+=nontvrooms*r2 (This should occur when -> no. of people == no. of non-tv rooms)
1 Like
I did what was exactly asked in the question and I got AC.
#include<bits/stdc++.h>
using namespace std;
#define ll long long int
#define lf long double
#define vi vector<int>
#define vl vector<long long int>
#define pii pair <int,int>
#define pll pair <long long int, long long int>
#define vpii vector< pair<int,int> >
#define vpll vector < pair <long long int,long long int> >
#define MOD 1000000007
#define pb push_back
#define mk make_pair
#define f first
#define s second
#define in insert
#define all(x) x.begin(),x.end()
#define zapp ios::sync_with_stdio(false);cin.tie(0)
const int MAXN = 1e4 + 5;
vector<bool> prime(MAXN, true);
void sieve()
{
prime[0] = prime[1] = false;
int i, j;
for (i = 2; i * i < MAXN; i++)
{
if (prime[i])
{
for (j = i * i; j <= MAXN; j += i)
prime[j] = false;
}
}
}
int main()
{
zapp;
sieve();
ll n1, n2;
cin >> n1 >> n2;
ll i, j = 0;
vector<string> bet;
for (i = n1; i <= n2; i++)
{
if (prime[i])
bet.pb(to_string(i));
}
vector<ll> nums;
ll n = bet.size();
for (i = 0; i < n; i++)
{
for (j = i + 1; j < n; j++)
{
string s = "";
s += bet[i];
s += bet[j];
stringstream geek(s);
ll x = 0;
geek >> x;
nums.pb(x);
string v = "";
v += bet[j];
v += bet[i];
stringstream meek(v);
ll y = 0;
meek >> y;
nums.pb(y);
}
}
sort(nums.begin(), nums.end());
set<ll> prm;
ll m = nums.size();
for (i = 0; i < m; i++)
{
if (prime[nums[i]])
prm.in(nums[i]);
}
ll cnt = prm.size();
vl act;
for (auto x : prm)
act.pb(x);
ll a = act[0];
ll b = act[cnt - 1];
ll fib[cnt];
fib[0] = a;
fib[1] = b;
for (i = 2; i < cnt; i++)
{
fib[i] = fib[i - 1] + fib[i - 2];
}
cout << fib[cnt - 1];
return 0;
}
If anyone here was able to solve E - Question, kindly share your approach 
if by E you mean lazy student then probability=1 (if number of unknowns n-m <number of questions in test paper t) or 1-[(n-m)Ct / nCt]
could you solve F btw? if yes…do share your approach
Damn, I used the same formula but then maybe fucked up in the nCr part. 
No man, I couldn’t implement it properly. It had something to do with the points’ images on the x-axis, but couldn’t figure it out.
1 Like
yeah i had a lot of trouble debugging that part too 
never mind…thanks anyways
My solution for E. Please debug it. Thanks
#include<bits/stdc++.h>
using namespace std;
#define ll long long int
#define lf long double
#define vi vector<int>
#define vl vector<long long int>
#define pii pair <int,int>
#define pll pair <long long int, long long int>
#define vpii vector< pair<int,int> >
#define vpll vector < pair <long long int,long long int> >
#define MOD 1000000007
#define pb push_back
#define mk make_pair
#define f first
#define s second
#define in insert
#define all(x) x.begin(),x.end()
#define zapp ios::sync_with_stdio(false);cin.tie(0)
ll add(ll x, ll y) {ll res = x + y; return (res >= MOD ? res - MOD : res);}
ll mul(ll x, ll y) {ll res = x * y; return (res >= MOD ? res % MOD : res);}
ll sub(ll x, ll y) {ll res = x - y; return (res < 0 ? res + MOD : res);}
ll power(ll x, ll y) {if (y < 0) return 1; ll res = 1; x %= MOD; while (y) {if (y & 1)res = mul(res, x); y >>= 1; x = mul(x, x);} return res;}
ll inv(ll x) {return power(x, MOD - 2);}
const ll mx = 1005;
ll fac[mx + 1];
void pre()
{
fac[0] = 1;
for (ll i = 1 ; i <= mx; i++)
fac[i] = fac[i - 1] * i % MOD;
}
ll nCr(ll n, ll r)
{
if (r == 0)
return 1;
return (fac[n] * inv(fac[r]) % MOD * inv(fac[n - r]) % MOD) % MOD;
}
int main()
{
pre();
ll t;
cin >> t;
while (t--)
{
ll n, b, m;
cin >> n >> b >> m;
ll q = nCr(n, b);
ll p;
if (m + b > n)
p = q;
else
p = nCr(n, b) - nCr(n - m, b);
ll g = __gcd(p, q);
p = p / g;
q = q / g;
cout << p*inv(q) << "\n";
}
return 0;
}
def fib(a,b,n):
sum=0
if n==1:
return a
elif n==2:
return b
else:
for i in range(n-2):
sum=a+b
a=b
b=sum
return sum
def cprim(a):
count=0
for i in range(1,a+1):
if a%i==0:
count+=1
if count==2:
return 1
else:
return 0
def ran(a,b):
li=[]
for i in range(a,b+1):
if(cprim(i)==1):
li.append(str(i))
return li
def merg(a):
nli=[]
for i in range(len(a)):
for j in range(len(a)):
if(i!=j):
b=int(a[i]+a[j])
if b not in nli:
nli.append(b)
nli.sort()
return nli
a,b=[int(x) for x in input().split()]
s1=ran(a,b)
s2=merg(s1)
s3=[]
for i in range(len(s2)):
if(cprim(s2[i])==1):
s3.append(s2[i])
ans=fib(s3[0],s3[-1],len(s3))
print(ans)