ZCMU训练赛-H（模拟）

H - Hard to Play

 

MightyHorse is playing a music game called osu!.

 

After playing for several months, MightyHorsediscovered the way of calculating score in osu!:

1. While playing osu!, player need to click some circles following the rhythm. Each time a player clicks, it will have three different points: 300, 100 and 50, deciding by how clicking timing fits the music.

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

 

P = Point * (Combo * 2 + 1)

 

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

Recently MightyHorse meets a high-end osu!player. After watching his replay, MightyHorsefinds that the game is very hard to play. But he is more interested in another problem: What's the maximum and minimum total score a player can get if he only knows the number of 300, 100 and 50 points the player gets in one play?

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

Input

There are multiple test cases.

The first line of input is an integer T (1 ≤ T≤ 100), indicating the number of test cases.

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

Output

For each test case, output a line contains two integers, describing the minimum and maximum total score the player can get.

Sample Input

1
2 1 1 

Sample Output

2050 3950

#include<stdio.h>
#include<ctype.h>
#include<string.h>
int main()
{
    int t;
    int a,b,c;
    int p1,p2,p3,p4,p5,p6;
    int mins,maxs;
    while(scanf("%d",&t)!=EOF)
    {
        mins=0;
        maxs=0;
        while(t--)
        {
           /* mins=0;
            maxs=0;*/
            p1=p2=p3=p4=p5=p6=0;
            scanf("%d%d%d",&a,&b,&c);
            for(int i=0;i<a;i++)
            {
                 p1+=300*(i*2+1);
            }
            for(int i=a;i<a+b;i++)
            {
                 p2+=100*(i*2+1);
            }
            for(int i=a+b;i<a+b+c;i++)
            {
                 p3+=50*(i*2+1);
            }
             mins=p1+p2+p3;

            for(int j=0;j<c;j++)
            {
                p4+=50*(j*2+1);
            }
            for(int j=c;j<c+b;j++)
            {
                 p5+=100*(j*2+1);
            }
            for(int j=c+b;j<a+b+c;j++)
            {
                 p6+=300*(j*2+1);
            }
             maxs=p4+p5+p6;
             printf("%d %d\n",mins,maxs);
        }

    }
    return 0;
}