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 <var>P</var> points, the total score will add <var>P</var>, which should be calculated according to following formula:

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

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 <var>T</var> (1 ≤ <var>T</var>≤ 100), indicating the number of test cases.

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

 1 #include<stdio.h>
 2 #include<ctype.h>
 3 #include<string.h>
 4 int main()
 5 {
 6     int t;
 7     int a,b,c;
 8     int p1,p2,p3,p4,p5,p6;
 9     int mins,maxs;
10     while(scanf("%d",&t)!=EOF)
11     {
12         mins=0;
13         maxs=0;
14         while(t--)
15         {
16            /* mins=0;
17             maxs=0;*/
18             p1=p2=p3=p4=p5=p6=0;
19             scanf("%d%d%d",&a,&b,&c);
20             for(int i=0;i<a;i++)
21             {
22                  p1+=300*(i*2+1);
23             }
24             for(int i=a;i<a+b;i++)
25             {
26                  p2+=100*(i*2+1);
27             }
28             for(int i=a+b;i<a+b+c;i++)
29             {
30                  p3+=50*(i*2+1);
31             }
32              mins=p1+p2+p3;
33 
34             for(int j=0;j<c;j++)
35             {
36                 p4+=50*(j*2+1);
37             }
38             for(int j=c;j<c+b;j++)
39             {
40                  p5+=100*(j*2+1);
41             }
42             for(int j=c+b;j<a+b+c;j++)
43             {
44                  p6+=300*(j*2+1);
45             }
46              maxs=p4+p5+p6;
47              printf("%d %d\n",mins,maxs);
48         }
49 
50     }
51     return 0;
52 }

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