Not able to find the issue for my solution to MEXUM (april lunchtime)

general
1

#include
#include
#include
#include
using namespace std;

typedef long l ;
typedef vector vi;
typedef vector vl;
template<typename… T>
void read(T&… args){
((cin>>args), …);
}

#define ford(i,n) for(int i=0;i<n;i++)
#define fore(i,n,m) for(int i=n;i<=m;i++)

#define pb push_back
#define all(x) x.begin(),x.end()

#define deb(x) cout<<#x<<" "<<x<<’\n’;
#define maxi 998244353;

l pow(int a, int b){
if(b==0)
return 1;
return (a*pow(a,b-1))%maxi;
}

int main(){
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int t;
read(t);
ford(q,t){
int n;
read(n);
vl a;
ford(i,n){
l temp;
read(temp);
a.pb(temp);
}
sort(all(a));
int k=a[0];
ford(i,n-1){
if(a[i]!=a[i+1]){
if(a[i]+1==a[i+1])
k = a[i+1];
else{
k = a[i];
break;
}
} else {
k = a[i];
}
}
if(a[0] != 1){
cout<<pow(2,n)<<’\n’;
continue;
}
if(k==1 && a[n-1]==1){
cout<<pow(2,n+1)-1<<’\n’;
continue;
}
vl frq;
int ptr=0;

int fcount=1;

while(a[ptr] <= k && ptr-1<n){
  if(a[ptr]!=a[ptr+1]){
      frq.pb(fcount);
      fcount=1;
  } else {
    fcount++;
  }
  ptr++;
}

l temp = 1;
l count = frq[0];;
l sum = pow(2,n-count);
  fore(i,2,k){
    temp = (temp*(pow(2,frq[i-2])-1))%maxi;
    count += frq[i-1];
    sum += (i * temp * pow(2,n-count))%maxi;
  }

temp = (temp*(pow(2,frq[k-1])-1))%maxi;
sum += ((k+1)*temp*pow(2,n-count))%maxi;
cout<<sum<<"\n";

}
return 0;
}

Thank you for your time.

2

Your solution would be giving TLE I suppose.

3

Your solution has a complexity of O(n^2) since you are finding power in linear time.
Read about binary exponentiation, it calculates power in logarithmic time.

4

Yes, the second case was coming TLE, but the first case came out to be WA. I am pretty sure the logic matches with the ones given by others in discussion but I am not able to figure out where my code fails.

5

Will definitely do that, thank you