[ACM_水题] ZOJ 3712 [Hard to Play 300 100 50 最大最小]

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

题目大意：一共有三种数字300，100，50，每个样例中给出数字的个数，公式是P = Point * (Combo * 2 + 1)，其中combo是这个数是第几个计算的。求P和的最大值和最小值。

解题思路：先算小的后算大的得出最大值，先算大的后算小的得出最小值

 #include<iostream>
 using namespace std;
 int main(){
     int T;
     cin>>T;
     while(T--){
         int A,B,C;
         cin>>A>>B>>C;
         int minSum=,maxSum=;
         int a=A,b=B,c=C;
         int cases=;//正着算最大值
         while(c--){
             maxSum+=cases*;
             cases+=;
         }
         while(b--){
             maxSum+=cases*;
             cases+=;
         }
         while(a--){
             maxSum+=cases*;
             cases+=;
         }
         cases=;//倒着算最小值
         while(A--){
             minSum+=cases*;
             cases+=;
         }
         while(B--){
             minSum+=cases*;
             cases+=;
         }
         while(C--){
             minSum+=cases*;
             cases+=;
         }
         cout<<minSum<<' '<<maxSum<<'\n';
     }return ;

 }