@gulshan_yadav
while(n–)
because n inputs will be given.
You made it too complex ig.You can check my solution here(not very clear but understandable)
number_of_testcases
{
int n ; cin >> n ;
int ans = 0 ;
string a ; cin >> a ;
while(n--){
string b ; cin >> b ;
if(b == "CONTEST_WON"){
ans += 300 ;
int rank ; cin >> rank;
if(rank <= 20)
ans +=20-rank;
}
else if(b == "TOP_CONTRIBUTOR"){
ans+=300;
}
else if(b == "BUG_FOUND"){
int severity ; cin >> severity ;
ans+=severity;
}
else if(b == "CONTEST_HOSTED"){
ans+=50;
}
}
(a == "INDIAN") ? (cout << ans/200) : (cout << ans/400) ;
cout << "\n";
}
2 Likes
This is my solution to coin flip and its shows TLE as complexity is O(n2). Is there any other way to optimise it?
#include <iostream>
using namespace std;
#define datatype long long int
int main()
{
datatype testcases;
cin>>testcases;
while(testcases--)
{
datatype games;
cin>>games;
while(games--)
{
datatype initialState, numberOfRounds, prediction;
cin>>initialState>>numberOfRounds>>prediction;
datatype arr[numberOfRounds];
if(initialState == 1) // head
for(datatype i=0; i<numberOfRounds; i++)
arr[i]=1;
else // tail
for(datatype i=0; i<numberOfRounds; i++)
arr[i]=-1;
for(datatype i=0; i<numberOfRounds; i++)
{
for(datatype j=0; j<i; j++)
arr[j]= -arr[j];
}
datatype count=0;
for(datatype i=0; i<numberOfRounds; i++)
{
if(prediction == 1)
{
if(arr[i] == -1)
count++;
}
else
{
if(arr[i] == 1)
count++;
}
}
cout<<count<<endl;
}
}
return 0;
}
1 Like
thanks you 
lost_boy12
i was taking string and rank both in one string only so that’s why its too complex.
3 Likes
for ZCO14003 iam getting AC for subtask 10,11,12 and 21 remaining are WA .I didn’t understand this.
1 Like
3
1
10
3
8 3 6
5
4 5 1 2 3
So for the 3rd test case answer should be 3 but here it is given 2 in output.
Can u please explain it??
1 Like
Why is that? Since 8 3 6 are in the order they entered lane. 8 can be at its max speed so does the 3. Only 6 cannot go to its peak speed as 3 is in front of it.
2 Likes
5 cannot be speeder than 4 and in the same way 2 & 3 cannot be speeder 1. So only 4 and 1 are at there max speed.
1 Like
consider all as bars of given size and every bar must be the smallest of all the bars in it’s left.
1 Like
BigInteger k = sc.nextBigInteger();
long d0 = sc.nextLong();
long d1 = sc.nextLong();
long sumofdigi = d1+d0;
long sum = d1+d0;
for(int i=2;i<k.intValue();i++){
long digi = sum%10;
sumofdigi+=digi;
sum+=digi;
}
if(sumofdigi%3==0)
System.out.println("YES");
else
System.out.println("NO");
it is showing TLE error but the complexity is O(n). Can anyone help?
1 Like
why i m getting wrong answer in LAPIN.i m getting right answer in my compiler for different test cases. plz explain
1 Like
Why is this approach incorrect?
1 Like
check constraints o(n) will not work
1 Like
For laddu problem, iterate for number of activities and everytime check the input. It should be one among 4 i.e, CONTEST_WON, TOP_CONTRIBUTOR, BUG_FOUND, CONTEST_HOSTED for a specific activity update the global variable using the given formulae. after iterating number of activities. divide global variable by 200 or 400 based on origin and then print it.
2 Likes
tell more about your approach.
1 Like
Because if you are taking the mod to each digits and then adding them, at the end you will have a number which is more than divider(i.e, 10 in the problem). so after that you will have to take the mod again.
2 Likes
Broad intuition is:
Lets say the number comes out to be 453, here 453 is divisible by 3, because (4 + 5 + 3) mod 3 = 0, but ((4 + 5 + 3) mod 10) mod 3 = (12 mod 10) mod 3 = 2 mod 3 = 2, so if you get the modulo of everything under 10, then it doesnt work, mainly because of the following statement:
((a mod 10 + b mod 10) mod 3) != (((a + b) mod 10) mod 3), for all a, b in [0, 9] * [0, 9]
2 Likes
int main() {
// your code goes here
ios_base::sync_with_stdio(false);
cin.tie(NULL);
long long n,fixprice=0;
cin >>n;
long long a[n];
unsigned long long sum[n]={0};
long long count[n]={0};
for(int i=0;i<n;i++)
{
cin >>a[i];
}
int size1=sizeof a/sizeof a[0];
sort(a, a+size1);
for(int j=0;j<n;j++)
{ fixprice=a[j];
count[j]=(n-j);
sum[j]=count[j]*fixprice;
}
int size=sizeof sum/sizeof sum[0];
sort(sum, sum+size, greater<int>());
cout << sum[0] <<"\n";
return 0;
}
This is the code