## Algorithm

Problem Name: 2 AD-HOC - beecrowd | 1836

# Pokémon!

Por Edson Alves, Faculdade UnB Gama Brazil

Timelimit: 1

In the first generation, the creatures from the Pokémon game have four basic attributes: hit points (HP), attack (AT), defense (DF) and speed (SP). These attributes increased as the pokémon gained experience levels winning battles or using special itens.

For each level (from 1 to 99) the value of the attributes can be computed by the following expressions:

e

where BS is the base value of the attribute (hit points, attack, defense and speed), EV is the effort value of the pokémon (it depends on what and how many battles the pokémon entered), IV is the individual value of the pokémon in the given attribute (it is the pokémon "gene") and L is the level.

The EV and IV values differentiate two pokémons of same kind, making both evolutions differ. Each attribute must have a integer value: the decimal value must be discarded after the fraction computation.

Given a pokémon, its basic attributes values and level, compute the attributes values according to the expressions.

## Input

The input consists of a series of test cases. The number of test cases T (T ≤ 1.000) is given in the first line of the input.

Each test case is composed by five lines. The first one contains the pokémon name P and its level L (1 ≤ L ≤ 99), separated by a single space. The pokémon's name has only alphanumeric characteres.

The following lines have three integers each: BS (1 ≤ BS ≤ 255), IV (1 ≤ IV ≤ 15) e EV (1 ≤ EV ≤ 262.140), separated by a single space, for each one of the attributes, in this order: HP, AT, DF e SP.

## Output

For each test case the output is formed by five messages, one per line:

1. Caso #t: P nivel L
2. HP: HPC
3. AT: ATC
4. DF: DFC
5. SP: SPC
where P is the pokémon's name, L the level and SC the attribute value for level L, according to given expressions, and t is the test case number.

 Input Samples Output Samples 4 Pikachu 81 35 7 22850 55 8 23140 30 13 17280 90 5 24795 Bulbasaur 50 45 9 20000 49 12 40000 49 3 60000 45 8 10000 Charmander 30 39 5 35000 52 14 60000 43 7 38000 65 15 200000 Squirtle 90 44 10 180000 48 2 220000 65 11 175000 43 8 192000 Caso #1: Pikachu nivel 81 HP: 189 AT: 137 DF: 101 SP: 190 Caso #2: Bulbasaur nivel 50 HP: 131 AT: 91 DF: 87 SP: 70 Caso #3: Charmander nivel 30 HP: 80 AT: 62 DF: 49 SP: 86 Caso #4: Squirtle nivel 90 HP: 292 AT: 200 DF: 235 SP: 195

## Code Examples

### #1 Code Example with C Programming

```Code - C Programming```

``````
#include <stdio.h>
#include <math.h>

int calcHP(unsigned nivel, unsigned base, unsigned gene, unsigned batalhas);
int calcS(unsigned nivel, unsigned base, unsigned gene, unsigned batalhas);

void main ()
{

unsigned gene, batalhas, base, nivel;
unsigned short casos;
char pokemon[1000], qtsCasos = 1;

scanf("%hu", &casos);

while (casos--)
{

scanf(" %s %u %u %u %u", pokemon, &nivel, &base, &gene, &batalhas);

printf("Caso #%hu: %s nivel %u\n", qtsCasos++, pokemon, nivel);
printf("HP: %d\n", calcHP(nivel, base, gene, batalhas));

scanf("%u %u %u", &base, &gene, &batalhas);
printf("AT: %d\n", calcS(nivel, base, gene, batalhas));

scanf("%u %u %u", &base, &gene, &batalhas);
printf("DF: %d\n", calcS(nivel, base, gene, batalhas));

scanf("%u %u %u", &base, &gene, &batalhas);
printf("SP: %d\n", calcS(nivel, base, gene, batalhas));

}

}

int calcHP(unsigned nivel, unsigned base, unsigned gene, unsigned batalhas)
{

double s;

s = (((gene + base + (sqrt(batalhas)/8) + 50) * nivel)/50) + 10;

return (int)s;

}

int calcS(unsigned nivel, unsigned base, unsigned gene, unsigned batalhas)
{

double s;

s = (((gene + base + (sqrt(batalhas)/8)) * nivel)/50) + 5;

return (int)s;

}
``````
Copy The Code &

Input

cmd
4
Pikachu 81
35 7 22850
55 8 23140
30 13 17280
90 5 24795
Bulbasaur 50
45 9 20000
49 12 40000
49 3 60000
45 8 10000
Charmander 30
39 5 35000
52 14 60000
43 7 38000
65 15 200000
Squirtle 90
44 10 180000
48 2 220000
65 11 175000
43 8 192000

Output

cmd
Caso #1: Pikachu nivel 81
HP: 189
AT: 137
DF: 101
SP: 190
Caso #2: Bulbasaur nivel 50
HP: 131
AT: 91
DF: 87
SP: 70 Caso #3: Charmander nivel 30
HP: 80
AT: 62
DF: 49
SP: 86
Caso #4: Squirtle nivel 90
HP: 292
AT: 200
DF: 235
SP: 195

### #2 Code Example with C++ Programming

```Code - C++ Programming```

``````
#include <bits/stdc++.h>

using namespace std;

int main()
{
string pk;
int n;
int l;
int bs, ev, iv;
int caso = 1;
cin >> n;

while (n--)
{
cin >> pk >> l;

cout << "Caso #" << caso++ << ": ";
cout << pk << " nivel " << l << '\n';
cin >> bs >> iv >> ev;
cout << "HP: " << (int)((((iv + bs + sqrt(ev)/8. + 50) * l) / 50) + 10) << '\n';
cin >> bs >> iv >> ev;
cout << "AT: " << (int)((((iv + bs + sqrt(ev)/8.) * l) / 50) + 5) << '\n';
cin >> bs >> iv >> ev;
cout << "DF: " << (int)((((iv + bs + sqrt(ev)/8.) * l) / 50) + 5) << '\n';
cin >> bs >> iv >> ev;
cout << "SP: " << (int)((((iv + bs + sqrt(ev)/8.) * l) / 50) + 5) << '\n';

}
}
``````
Copy The Code &

Input

cmd
4
Pikachu 81
35 7 22850
55 8 23140
30 13 17280
90 5 24795
Bulbasaur 50
45 9 20000
49 12 40000
49 3 60000
45 8 10000
Charmander 30
39 5 35000
52 14 60000
43 7 38000
65 15 200000
Squirtle 90
44 10 180000
48 2 220000
65 11 175000
43 8 192000

Output

cmd
Caso #1: Pikachu nivel 81
HP: 189
AT: 137
DF: 101
SP: 190
Caso #2: Bulbasaur nivel 50
HP: 131
AT: 91
DF: 87
SP: 70 Caso #3: Charmander nivel 30
HP: 80
AT: 62
DF: 49
SP: 86
Caso #4: Squirtle nivel 90
HP: 292
AT: 200
DF: 235
SP: 195