ZOJ 3712 Hard to Play

Hard to Play

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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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 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 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

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Author: DAI, Longao

Contest: The 10th Zhejiang Provincial Collegiate Programming Contest

數學題目，簡單模擬

#include<stdio.h>
using namespace std;
int main()
{
  int t;
  scanf("%d",&t);
  while(t--)
  {
    int a,b,c;
    scanf("%d%d%d",&a,&b,&c);
    int min=0,max=0;
    max=50*(c*c)+100*((b+c)*(b+c)-c*c)+300*((a+b+c)*(a+b+c)-(b+c)*(b+c));
    min=300*(a*a)+100*((a+b)*(a+b)-a*a)+50*((a+b+c)*(a+b+c)-(a+b)*(a+b));
    printf("%d %d\n",min,max);
  }
  return 0;
}

簡單題解：

由上圖題目中所給的公式，可推出（Combo*2+1）通項公式爲n*n。

樣例解釋：

max：

50*（0*2+1）=50

100*（1*2+1）=300

300*（2*2+1）=1500

300*（3*2+1）=2100

min：

300*（0*2+1）=300

300*（1*2+1）=900

100*（2*2+1）=500

50*（3*2+1）=350

又上面所容易得出：

（0*2+1）=1；      

（0*2+1）+（1*2+1）=4；

（0*2+1）+（1*2+1）+（2*2+1）=9；

（0*2+1）+（1*2+1）+（2*2+1）+（3*2+1）=16；

則可得出通項公式n*n。