1884: 三个家庭（思维题）

1884: 三个家庭
Time Limit: 1 Sec Memory Limit: 128 MB
Submit: 244 Solved: 81
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Description

There are three families who have a garden, each wife of three families need to help clean the garden. Mrs. A worked for five days and Mrs. B worked for four days before finishing the garden. Mis. C is pregnant so she can’t join their work.In order to thank for their work,she paid 90 yuan .The question is how to give A,B the money , how much should A get? 90 / (5 + 4) * 5 = 50 yuan ? If you think so , how stupid you are ! The correct answer is 60 yuan . Please think about it carefully.
The following answer is a general question: assuming that Mrs. A worked for x days, Mrs. B worked for y days, Mrs. C had z yuan , how much should A wife get ? Two wives get non-negative integer money. Three wives have the same work efficiency.

Input

Multiple cases , for each case:
Three positive integers x,y,z . A wife worked for x days , B wife worked for y days , and C wife paid z yuan for them .

Output

For each case ,output a interger ,the money that A wife got. We needed to do decimals to round up and round down numbers。

Sample Input

5 4 90

Sample Output

HINT

Source

XF
AC_code:
/*
思维题，本来三个人应该做一样多的，现在A,B帮C做了她那份，看A帮C做了多少，注意这句话： We needed to do decimals to round up and round down numbers（我们需要做小数的四舍五入）。
*/

#include <stdio.h>
int main()
{double x,y,z;while(~scanf("%lf%lf%lf",&x,&y,&z)){double aver = (x + y)/3;double A_get;x > aver?A_get = (x - aver)*z/aver:A_get = 0;printf("%.0lf\n",A_get);}return 0;
}

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