Three Families

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=3946

Three Families

Three families share a garden. They usually clean the garden together at the end of each week, but last week, family C was on holiday, so family A spent 5 hours, family B spent 4 hours and had everything done. After coming back, family C is willing to pay $90 to the other two families. How much should family A get? You may assume both families were cleaning at the same speed.

$90/(5+4)*5=$50? No no no. Think hard. The correct answer is $60. When you figured out why, answer the following question: If family A and B spent x and y hours respectively, and family C paid $z, how much should family A get? It is guaranteed that both families should get non-negative integer dollars.

WARNING: Try to avoid floating-point numbers. If you really need to, be careful!

Input

The first line contains an integer 
T
 (
T100
), the number of test cases. Each test case contains three integers 
x
, 
y
, 
z
 (
1x
, 
y10
,

1z1000
).

Output

For each test case, print an integer, representing the amount of dollars that family A should get.

Sample Input

2
5 4 90
8 4 123

Sample Output

60
123

---

Problemsetter: Rujia Liu, Special Thanks: Feng Chen, Md. Mahbubul Hasann, Youzhi Bao

这个题就以纯粹数学题，能想通公式则不要10行代码解决，想不到公式想死都做不出来。。。

这题是2012年湖南省省赛签到题。。。。。

思路：他们三家共用一个花园，还是平局分的，三家工作效率相同，那么也就是说三家修剪草坪的总时间相同，现在只有2家人做3家人的活。

则每家修剪草坪的时间为： 
(x+y)/3

由此可知A，B两家给C修剪的时间分别为：x-(x+y)/3   和  y-(x+y)/3

C给A、B支付报酬是按比例支付的，则A的报酬为：z*(x-(x+y)/3)/(x-(x+y)/3+y-(x+y)/3)

AC代码：

#include<iostream>
#include<cstdio>

int main()
{
    int t,x,y,z,mu;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d%d",&x,&y,&z);
        mu = z*(x-(x+y)/3)/(x-(x+y)/3+y-(x+y)/3);
        printf("%d\n",mu);
    }

    return 0;
}