zoj 3712 Hard to Play（数学题）

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题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3712

题意：

分别输出得分最大值和最小值！Combo开始为0，随后一直加1，知道结束！因为MightyHorse从不失手！

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 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

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Author:  DAI, Longao
Contest:  The 10th Zhejiang Provincial Collegiate Programming Contest

代码如下：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
using namespace std;
int main()
{
	int i,j,m,n,t,sum,s,max,x,min;
	int a,b,c;
	while(scanf("%d",&t)!=EOF)
	{
		while(t--)
		{
			min = max = 0;
			scanf("%d%d%d",&a,&b,&c);
			s = a+b+c;
			for(j = 0 ;j < a ; j++ )
				min += 300 *(2 * j + 1);
			for(j = a ; j<a+b ;j++)
			{
				min+=100*(2*j+1);
			}
			for(j = a+b; j < s ; j++ )
			{
				min+=50*(2 * j + 1);
			}
			for(j = 0 ;j < c ; j++ )
				max += 50 *(2 * j + 1);
			for(j = c ; j < c+b ;j++)
			{
				max+=100*(2*j+1);
			}
			for(j = b+c; j < s ; j++ )
			{
				max+=300*(2 * j + 1);
			}
			printf("%d %d\n",min,max);
		}
	}
	return 0;
}