ZOJ 3712 Hard to Play

Description

MightyHorse isplaying a music game called osu!.

Afterplaying for several months, MightyHorse discoveredthe way of calculating score in osu!:

1. Whileplaying osu!, player need toclick some circles following the rhythm. Each time a player clicks, it willhave three different points: 300, 100 and 50, deciding by how clicking timingfits the music.

2.Calculating the score is quite simple. Each time player clicks and gets P points, the totalscore will add P, whichshould be calculated according to following formula:

P = Point * (Combo * 2 + 1)

Here Point is the point theplayer gets (300, 100 or 50) and Combo is thenumber of consecutive circles the player gets points previously - That means ifthe player doesn't miss any circle and clicks the i^th circle, Combo shouldbe i - 1.

Recently MightyHorse meets a high-end osu! player. Afterwatching his replay, MightyHorse findsthat the game is very hard to play. But he is more interested in anotherproblem: What's the maximum and minimum total score a player can get if he onlyknows the number of 300, 100 and 50 points the player gets in one play?

As thehigh-end player plays so well, we can assume that he won't miss any circlewhile playing osu!; Thushe can get at least 50 point for a circle.

Input

Thereare multiple test cases.

Thefirst line of input is an integer T (1 ≤ T ≤ 100), indicatingthe number of test cases.

For eachtest case, there is only one line contains three integers: A (0 ≤ A ≤ 500) - the numberof 300 point he gets, B (0 ≤ B ≤ 500) - the numberof 100 point he gets and C (0 ≤ C ≤ 500) - the numberof 50 point he gets.

Output

For eachtest case, output a line contains two integers, describing the minimum andmaximum total score the player can get.

Sample Input

1

2 1 1

Sample Output

2050 3950

#include <iostream>

#include <string.h>

using namespace std;

int main()

{

   int T;

   cin>>T;

   while(T--)

    {

      int A,B,C;

      cin>>A>>B>>C;

      long long mins=0,maxs=0;

      maxs=maxs+50*C*(C-1)+50*C;

      maxs=maxs+100*B*(2*C+B-1)+100*B;

      maxs=maxs+300*A*(A+2*B+2*C-1)+300*A;

      mins=mins+300*A*(A-1)+300*A;

      mins=mins+100*B*(2*A+B-1)+100*B;

      mins=mins+50*C*(2*A+2*B+C-1)+50*C;

      cout<<mins<<" "<<maxs<<endl;

    }

       return 0;

}