HackerEarth Optimization Game problem solution

In this HackerEarth Optimization Game problem solution Currently, Monk is playing a unique kind of strategy game called an optimization game. In this game, we are provided with an array containing integral numbers. Now all these numbers represent the count of their respective index power of 2. The goal of the game is to minimize the total sum of the count of the array by converting lower powers of 2 into their higher powers i.e.

HackerEarth Optimization Game problem solution

HackerEarth Optimization Game problem solution.

using namespace std;

#define TRACE
#ifdef TRACE
#define TR(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
cerr << name << " : " << arg1 << std::endl;
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
const char* comma = strchr(names + 1, ',');cerr.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
#define TR(...)

typedef long long LL;
typedef pair<int,int> II;
typedef vector<int> VI;
typedef vector<II> VII;

#define REP(i,i1,n) for(int i=i1;i<n;i++)
#define REPB(i,i1,n) for(int i=i1;i>=n;i--)
#define PB push_back
#define MP make_pair
#define ALL(c) (c).begin(),(c).end()
#define F first
#define S second
#define log2 0.30102999566398119521373889472449L
#define SZ(a) (int)a.size()
#define EPS 1e-12
#define MOD 1000000007
#define FAST_IO ios_base::sync_with_stdio(false);cin.tie(NULL)
#define SI(c) scanf("%d",&c)
#define SLL(c) scanf("%lld",&c)
#define PIN(c) printf("%d\n",c)
#define PLLN(c) printf("%lld\n",c)
#define N 100010
#define endl '\n'
#define FILL(ar,vl) for(int i=0;i<N;i++)ar[i]=vl
#define FILL2(ar,vl) for(int i=0;i<N;i++)for(int j=0;j<N;j++)ar[i][j]=vl

inline int mult(int a , int b) { LL x = a; x *= LL(b); if(x >= MOD) x %= MOD; return x; }
inline int add(int a , int b) { return a + b >= MOD ? a + b - MOD : a + b; }
LL powmod(LL a,LL b) { if(b==0)return 1; LL x=powmod(a,b/2); LL y=(x*x)%MOD; if(b%2) return (a*y)%MOD; return y%MOD; }

LL a[N + 200];
int main() {
int t;
while(t--) {
int n;
int ans = 0;
REP(i,0,N+200) a[i] = 0;
REP(i,0,n) cin>>a[i];
REP(i,0,N+200) {
a[i+1] += a[i] / 2;
a[i] %= 2;
ans += a[i];
return 0;

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