ZOJ 3712 Hard to Play

Hard to Play

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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MightyHorse is playing a music game called osu!.

After playing for several months, MightyHorse discovered 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, MightyHorse finds 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

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Author: DAI, Longao
Contest: The 10th Zhejiang Provincial Collegiate Programming Contest

 

 1 #include <cstdio>
 2 using namespace std;
 3 
 4 int main()
 5 {
 6     int ca, a, b, c,min,max,com;
 7     scanf("%d", &ca);
 8     while (ca--)
 9     {
10         scanf("%d%d%d", &a, &b, &c);
11         min = max = 0; com = 0;
12         for (int i = 0; i < a; i++, com++)
13             min  += 300 * (com * 2 + 1);
14         for (int i = 0; i < b; i++, com++)
15             min += 100 * (com * 2 + 1);
16         for (int i = 0; i < c; i++, com++)
17             min += 50 * (com * 2 + 1);
18         com = 0;
19         for (int i = 0; i < c; i++, com++)
20             max  += 50 * (com * 2 + 1);
21         for (int i = 0; i < b; i++, com++)
22             max += 100 * (com * 2 + 1);
23         for (int i = 0; i < a; i++, com++)
24             max += 300 * (com * 2 + 1);
25         printf("%d %d\n", min,max);
26     }
27 }