UVA 12502 Three Families （A）

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

思路：主要要考虑到A,B除了帮第三个家庭分担外，本身自己也是要做工作的。所以5,4共做9小时，那么每个家庭要做3个小时。
则A帮C分担了2个小时，B帮C分担了1个小时，所以A获得90*2/3.
得出公式后时除法可能出错，要放到最后。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <string>
using namespace std;
int x, y, z;
int T;
int main(){
    scanf("%d", &T);
    while(T--){
        scanf("%d%d%d", &x, &y, &z);
        if(x > (x+y)/3){
            printf("%d
",(2*x-y)*z/(x+y));
        }
        else printf("0
");
    }
    
    return 0;
}