ZOJ3712：Hard to Play
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:
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 ith circle, Comboshould 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
题意：一共有三种数字300，100，50，每个样例中给出数字的个数，公式是P = Point * (Combo * 2 + 1)，其中combo是这个数是第几个计算的，
结果中的两个，一个是从左到右计算，另一个是从又到左计算
#include <stdio.h>
#include <string.h>
int main()
{
int t;
int a[4] = {300,100,50};
int i,v[4],sum,x,cas;
scanf("%d",&t);
while(t--)
{
scanf("%d%d%d",&v[0],&v[1],&v[2]);
x = 1;
sum = 0;
for(i = 0; i<3; i++)
{
cas = v[i];
while(cas--)
{
sum+=a[i]*((x-1)*2+1);
x++;
}
}
printf("%d ",sum);
x = 1;
sum = 0;
for(i = 2; i>=0; i--)
{
cas = v[i];
while(cas--)
{
sum+=a[i]*((x-1)*2+1);
x++;
}
}
printf("%d\n",sum);
}
return 0;
}