https://www.codechef.com/viewsolution/32272069

This only passes the first task in Subtask 1 , which makes me believe my approach is correct but my implementation is wrong.

I calculate ans as,

ans = 1*( 2^{(cnt of elements \gt 1)} ) + 2*( 2^{(cnt of elements > 2)} ) * (2^{(number of 1s)}-1) + 3*( 2^{(cnt of elements > 3)} ) * (2^{(number of 1s)}-1) * (2^{(number of 2s)}-1) + .... until we find a number not present in the sequence.

```
#include <bits/stdc++.h>
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
typedef double db;
const ll mod = 998244353;
const db eps = 1e-9 ;
const ll maxn = 1e9+1;
const ll inf = LLONG_MAX;
#define mp make_pair
#define pb push_back
#define endl "\n"
#define deb(x) cout << #x << " " << x << endl
ll bin_power_mod(ll base, ll power, ll m)
{
base %= m;
ll res = 1;
while (power > 0) {
if (power & 1)
res = res * base % m;
base = base * base % m;
power >>= 1;
}
return res;
}
void solve()
{
ll n , ans = 0 , multi = 1 , total =0;
cin >> n;
set<ll> s;
map<ll,ll> cnt;
for(ll i=0,tmp ; i<n ; ++i)
{
cin >> tmp;
s.insert(tmp);
cnt[tmp]++;
}
for( ll i=1 ; i < maxn ; ++i)
{
if( s.find(i) != s.end() )
{
total += cnt[i];
ans += (i*( bin_power_mod( 2LL , n-total , mod)*multi)%mod)%mod;
multi *= ( bin_power_mod(2LL ,cnt[i] , mod)- 1) ;
ans%=mod , multi%=mod;
}
else
{
ans += (i*(bin_power_mod( 2 ,n-total , mod)*multi)%mod)%mod;
ans%=mod ;
cout << ans << endl;
return;
}
}
return;
}
int main()
{
ios_base :: sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
#ifndef ONLINE_JUDGE
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
freopen("error.txt","w",stderr);
#endif
ll t;
cin >> t;
while(t--)
{
solve();
}
return 0;
}
```