zoj-3712 Hard to Play

Hard to Play

---

Time Limit: 2 Seconds      Memory Limit: 65536 KB

---

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

//简单题，秒
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<iostream>
using namespace std;
int main()
{
    int  t;
    scanf("%d",&t);
    int num1,num2,num3;
    while(t--)
    {
        scanf("%d%d%d",&num1,&num2,&num3);
        int i,j,k;
        int mmax,mmin;
        mmax=mmin=0;
        for(i=0;i<num1;i++)
            mmin+=300*(i*2+1);
        for(j=i;j<num1+num2;j++)
            mmin+=100*(j*2+1);
        for(k=j;k<num1+num2+num3;k++)
            mmin+=50*(k*2+1);
        for(i=0;i<num3;i++)
            mmax+=50*(i*2+1);
        for(j=i;j<num3+num2;j++)
            mmax+=100*(j*2+1);
        for(k=j;k<num1+num2+num3;k++)
            mmax+=300*(k*2+1);
            printf("%d %d\n",mmin,mmax);
    }
    return 0;
}

转载于:https://www.cnblogs.com/zafuacm/archive/2013/05/20/3089486.html