2014ACM集训13级PK赛3-Hard to Play

Description

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

12 1 1 

Sample Output

2050 3950

 

有点像节奏大师，怪不得飞飞这么快就A了。

#include <stdio.h>#include <string.h>#include <stdlib.h>int main(){    int N;    scanf ("%d",&N);    while (N--)    {        int imax = 0,imin = 0;        int a,b,c;        int i;        scanf ("%d%d%d",&a,&b,&c);        int add = a + b + c;        for (i = 0;i < add;i++)        {            if(i < c)                imax += 50 * (i * 2 + 1);            else if (i >= c && i < c + b)                imax += 100 * (i * 2 + 1);            else                imax += 300 * (i * 2 + 1);            if (i < a)                imin += 300 * (i * 2 + 1);            else if (i >= a && i < b + a)                imin += 100 * (i * 2 + 1);            else                imin += 50 * (i * 2 + 1);        }        printf ("%d %d\n",imin,imax);    }    return 0;}