| 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 (T
100), 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。
1 #include <iostream> 2 #include <cstring> 3 #include <cstdio> 4 #include <string> 5 using namespace std; 6 int x, y, z; 7 int T; 8 int main(){ 9 scanf("%d", &T); 10 while(T--){ 11 scanf("%d%d%d", &x, &y, &z); 12 if(x > (x+y)/3){ 13 printf("%d ",(2*x-y)*z/(x+y)); 14 } 15 else printf("0 "); 16 } 17 18 return 0; 19 }