# PLRTRAVELS Editorial

Contest

Author: Amul Agarwal
Tester: Ishaan Shah, Sanyam Shah, Sambasai Andem and Tejas Chaudhari
Editorialist: Rutvij Menavlikar

EASY

# EXPLANATION:

In this problem, all the required hints were hidden in the sample test case.
For each planet, we define its base as the index of it when we write planets in increasing order of distance from the sun. For example, Mercury has base 1, Venus has base 2, Neptune has base 8.
For this problem, you had to do the base conversion of departure time to get the arrival time.

# SOLUTIONS:

Setter's Solution
``````planets = ["mercury", "venus", "earth", "mars",
"jupiter", "saturn", "uranus", "neptune"]

def tobase10(s, base):
if base == 1:
return len(str(s))
s = int(s)
ans, pw = 0, 1
while s > 0:
ans += pw * (s % 10)
s = s // 10
pw *= base
return ans

def tobasek(s, k):
s = int(s)
if k == 1:
return "1" * s
digits = []
while s:
digits.append(str(s % k))
s //= k
return "".join(digits[::-1])

T = int(input())
for _ in range(T):
dep, arr, tm = map(str, input().split())
from_base = planets.index(dep) + 1
to_base = planets.index(arr) + 1
base_10 = tobase10(tm, from_base)
base_to = tobasek(base_10, to_base)
print(base_to)
``````
Tester's Solution
``````#include <bits/stdc++.h>
using namespace std;
#define fast                      \
ios_base::sync_with_stdio(0); \
cin.tie(0);                   \
cout.tie(0)
#define tc       \
long long t; \
cin >> t;    \
while (t--)

long long retbase(string s)
{
if (s == "mercury")
{
return 1;
}
else if (s == "venus")
{
return 2;
}
else if (s == "earth")
{
return 3;
}
else if (s == "mars")
{
return 4;
}
else if (s == "jupiter")
{
return 5;
}
else if (s == "saturn")
{
return 6;
}
else if (s == "uranus")
{
return 7;
}
else if (s == "neptune")
{
return 8;
}
else if (s == "pluto")
{
return 9;
}
return -1;
}

// Function to return ASCII
// value of a character
long long val(char c)
{
if (c >= '0' && c <= '9')
return (long long)c - '0';
else
return (long long)c - 'a' + 10;
}

long long toDeci(string str, long long base)
{
// Stores the length
// of the string
long long len = str.size();

// Initialize power of base
long long power = 1;

// Initialize result
long long num = 0;

// Decimal equivalent is str[len-1]*1
// + str[len-2]*base + str[len-3]*(base^2) + ...
for (long long i = len - 1; i >= 0; i--)
{

// A digit in input number must
// be less than number's base
if (val(str[i]) > base)
{
cout << val(str[i]) << endl;
printf("Invalid Number");
return -1;
}

// Update num
num += val(str[i]) * power;

// Update power
power = power * base;
}

return num;
}

// Function to return equivalent
// character of a given value
char reVal(long long num)
{
if (num >= 0 && num <= 9)
return (char)(num + '0');
else
return (char)(num - 10 + 'A');
}

// Function to convert a given
// decimal number to a given base
string fromDeci(long long base, long long inputNum)
{
// Store the result
string res = "";
if (base == 1)
{
while (inputNum > 0)
{
res += "1";
inputNum--;
}
}
else
{
// Repeatedly divide inputNum
// by base and take remainder
while (inputNum > 0)
{

// Update res
res += reVal(inputNum % base);

// Update inputNum
inputNum /= base;
}
}

// Reverse the result
reverse(res.begin(), res.end());

return res;
}

// Function to convert a given number
// from a base to another base
void convertBase(string s, long long a, long long b)
{
// Convert the number from
// base A to decimal
long long num = toDeci(s, a);

// Convert the number from
// decimal to base B
string ans = fromDeci(b, num);

// Print the result
cout << ans << endl;
}

signed main()
{
fast;
tc
{
long long base1, base2;
string home, away, dep_time;
cin >> home >> away >> dep_time;
base1 = retbase(home);
base2 = retbase(away);
convertBase(dep_time, base1, base2);
}
}
``````
Editorialist's Solution
``````#include <bits/stdc++.h>

using namespace std;

#define ll long long
#define pb push_back
#define io                            \
ios_base::sync_with_stdio(false); \
cin.tie(NULL);                    \
cout.tie(NULL)
#define all(v) v.begin(), v.end()
#define tt     \
ll tc;     \
cin >> tc; \
while (tc--)
#define each(x, v) for (auto x : v)

ll ind(string s)
{
if (s == "mercury")
return 1;
else if (s == "venus")
return 2;
else if (s == "earth")
return 3;
else if (s == "mars")
return 4;
else if (s == "jupiter")
return 5;
else if (s == "saturn")
return 6;
else if (s == "uranus")
return 7;
else if (s == "neptune")
return 8;
else
return -1;
}

ll num(string s, ll b)
{
ll ans = 0, a = 1;
reverse(all(s));
each(c, s)
{
ans += a * (c - '0');
a *= b;
}
return ans;
}

string str(ll n, ll b)
{
string ans = "";
if (b == 1)
{
for (ll i = 0; i < n; ++i)
ans.pb('1');
return ans;
}
while (n)
{
ans.pb('0' + (n % b));
n /= b;
}
reverse(all(ans));
return ans;
}

int main()
{
io;
tt
{
string s1, s2, s3;
cin >> s1 >> s2 >> s3;
ll b1 = ind(s1), b2 = ind(s2);
cout << str(num(s3, b1), b2) << endl;
}
return 0;
}
``````