ZOJ 3712Hard to Play（水题）

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

12 1 1 

Sample Output

2050 3950

想法：水题

代码：
#include<stdio.h>#include<string.h>int main(){  int T;  scanf("%d",&T);  while(T--)  {    int a,b,c;    scanf("%d %d %d",&a,&b,&c);    int i,j;    long long  sum1=0;    for(i=1;i<=a+b+c;i++)    {        if(i<=a)        {            sum1+=(2*(i-1)+1)*300;        }        else        {            if(i<=a+b)            {              sum1+=(2*(i-1)+1)*100;            }            else            {              sum1+=(2*(i-1)+1)*50;            }        }    }    long long sum=0;    for(i=1;i<=a+b+c;i++)    {        if(i<=c)        {            sum+=(2*(i-1)+1)*50;        }        else        {            if(i<=c+b)            {              sum+=(2*(i-1)+1)*100;            }            else            {              sum+=(2*(i-1)+1)*300;            }        }    }    printf("%lld %lld\n",sum1,sum);  }  return 0;}