# DRUNKALK - Editorial

Author, Editorialist : Raghav Agarwal
Tester : Miten Shah

Cakewalk

None

# PROBLEM:

A person moves 3 steps forward in odd second and 1 step backward in even seconds. Calculate the distance traveled by the person after n seconds.

# EXPLANATION:

### Linear time

Since the constraints are small we could also simulate the whole walk. By incrementing distance by 3 for odd seconds and decreasing it by 1 in even seconds.

### Constant time

In 2 seconds the person moves 3 forward and 1 backward, that is he moves a net of 2 steps forward every two seconds. Thus if n is even the total distance travelled will be just n. While if n is odd, then n - 1 is even, and the distance travelled in n - 1 seconds is n - 1, then in the last second 3 more forward steps are taken. So total distance will be n - 1 + 3 = n + 2 for odd n.

# SOLUTION :

c++ Solution (Linear)
``````#include <bits/stdc++.h>
#define fast_io std::ios::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL)

using namespace std;

int main() {
fast_io;
long long t;
cin >> t;

int sum = 0;
while (t--) {
bool f = 1;
int k;
cin >> k;
sum += k;

long long ans = 0;

while (k--) {
ans += f ? 3 : -1;
f = !f;
}
cout << ans << '\n';
}
}
``````
C++ Tester's solution
``````// created by mtnshh

#include<bits/stdc++.h>
using namespace std;
#define ll long long int
#define pb push_back
#define rb pop_back
#define ti tuple<int, int, int>
#define pii pair<int, int>
#define pli pair<ll, int>
#define pll pair<ll, ll>
#define mp make_pair
#define mt make_tuple

#define rep(i,a,b) for(ll i=a;i<b;i++)
#define repb(i,a,b) for(ll i=a;i>=b;i--)

#define err() cout<<"--------------------------"<<endl;
#define errA(A) for(auto i:A)   cout<<i<<" ";cout<<endl;
#define err1(a) cout<<#a<<" "<<a<<endl
#define err2(a,b) cout<<#a<<" "<<a<<" "<<#b<<" "<<b<<endl
#define err3(a,b,c) cout<<#a<<" "<<a<<" "<<#b<<" "<<b<<" "<<#c<<" "<<c<<endl
#define err4(a,b,c,d) cout<<#a<<" "<<a<<" "<<#b<<" "<<b<<" "<<#c<<" "<<c<<" "<<#d<<" "<<d<<endl

#define all(A)  A.begin(),A.end()
#define allr(A)    A.rbegin(),A.rend()
#define ft first
#define sd second

#define V vector<ll>
#define S set<ll>
#define VV vector<V>
#define Vpll vector<pll>

#define endl "\n"

long long readInt(long long l,long long r,char endd){
long long x=0;
int cnt=0;
int fi=-1;
bool is_neg=false;
while(true){
char g=getchar();
// char g = getc(fp);
if(g=='-'){
assert(fi==-1);
is_neg=true;
continue;
}
if('0'<=g && g<='9'){
x*=10;
x+=g-'0';
if(cnt==0){
fi=g-'0';
}
cnt++;
assert(fi!=0 || cnt==1);
assert(fi!=0 || is_neg==false);

assert(!(cnt>19 || ( cnt==19 && fi>1) ));
} else if(g==endd){
if(is_neg){
x= -x;
}
// cerr << x << " " << l << " " << r << endl;
assert(l<=x && x<=r);
return x;
} else {
assert(false);
}
}
}
string readString(int l,int r,char endd){
string ret="";
int cnt=0;
while(true){
char g=getchar();
// char g=getc(fp);
assert(g != -1);
if(g==endd){
break;
}
cnt++;
ret+=g;
}
assert(l<=cnt && cnt<=r);
return ret;
}
long long readIntSp(long long l,long long r){
}
long long readIntLn(long long l,long long r){
}
string readStringLn(int l,int r){
}
string readStringSp(int l,int r){
}

const int max_q = 1e5;
const int max_k = 1e5;
const int max_sum_k = 1e6;

const ll N = 100005;
const ll INF = 1e12;

void solve(ll k){
if(k % 2 == 0){
cout << k << endl;
}
else{
cout << k + 2 << endl;
}
}

int main(){
ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
ll q = readIntLn(1, max_q);
ll sum_k = 0;
while(q--){
ll k = readIntLn(0, max_k);
solve(k);
sum_k += k;
}
assert(sum_k <= max_sum_k);
assert(getchar()==-1);
}
``````