[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

---

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=0,maxSum=0;
        int a=A,b=B,c=C;
        int cases=1;//正着算最大值
        while(c--){
            maxSum+=cases*50;
            cases+=2;
        }
        while(b--){
            maxSum+=cases*100;
            cases+=2;
        }
        while(a--){
            maxSum+=cases*300;
            cases+=2;
        }
        cases=1;//倒着算最小值
        while(A--){
            minSum+=cases*300;
            cases+=2;
        }
        while(B--){
            minSum+=cases*100;
            cases+=2;
        }
        while(C--){
            minSum+=cases*50;
            cases+=2;
        }
        cout<<minSum<<' '<<maxSum<<'
';
    }return 0;

}