浙江省ACM竞赛（2013）---H - Hard to Play

H - Hard to Play

Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu

Submit  Status  Practice  ZOJ 3712

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

1
2 1 1 

Sample Output

2050 3950

AC CODE：

#include <iostream>
using namespace std;
int main()
{
    int T;
    cin>>T;
    while(T--)
    {
        int n300,n100,n50;
        cin>>n300>>n100>>n50;
        int n=n300+n100+n50;
        int i;
        int k1=n300;
        int k2=n100;
        int k3=n50;
        int min=0;
        int max=0;
        for(i=0; i<n; i++)
        {
            if(k1>0)
            {
                k1--;
                min+=((i<<1)+1)*300;
                continue;
            }
            if(k2)
            {
                k2--;
                min+=((i<<1)+1)*100;
                continue;
            }
            if(k3)
            {
                k3--;
                min+=((i<<1)+1)*50;
                continue;
            }
        }
        k1=n50;
        k2=n100;
        k3=n300;
        for(i=0;i<n; i++)
        {
            if(k1>0)
            {
                k1--;
                max+=((i<<1)+1)*50;
                continue;
            }
            if(k2>0)
            {
                k2--;
                max+=((i<<1)+1)*100;
                continue;
            }
            if(k3>0)
            {
                k3--;
                max+=((i<<1)+1)*300;
                continue;
            }
        }
        cout<<min<<" "<<max<<endl;
    }
    return 0;
}