ZOJ-Hard to Play

题目

MightyHorse is playing a music game called osu!.

After playing for several months, 

MightyHorse discovered the way of calculating score inosu!:

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 thei^th circle, Combo should bei - 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 playingosu!; 

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 andC (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

#include <bits/stdc++.h>//（注意此题中的i<th>,由此可推得先加50分，再加100、300可得最大分值，反之得最小分值）
using namespace std;
int main()
{
    int T;
    while(scanf("%d",&T)!=EOF)
    {
        while(T--)
        {
            int a,b,c;
            scanf("%d%d%d",&a,&b,&c);
           int num=a+b+c;
           int i;
           int ma=0,mi=0;
           for(i=0;i<num;i++)
           {
               if(i<c)
               {
                   ma=ma+50*(i*2+1);
               }
               else if(i>=c&&i<b+c)
               {
                   ma=ma+100*(i*2+1);
               }
               else if(i>=b+c)
               {
                   ma=ma+300*(i*2+1);
               }
               if(i<a)
               {
                   mi=mi+300*(i*2+1);
               }
               else if(i>=a&&i<a+b)
               {
                   mi=mi+100*(i*2+1);
               }
               else if(i>=b+a)
               {
                   mi=mi+50*(i*2+1);
               }
           }
           printf("%d %d\n",mi,ma);
        }
    }
    return 0;
}