1111: Three tips for people to solve problems

Original question:

1111: Three family time limit:1 Sec Memory limit:128 MB
Description has three families with a garden, and the wives of each family are required to help organize the garden. Mrs. A worked for 5 days, and Mrs. B worked for 4 days before finishing the garden. Mrs. C was 90 yuan because she was pregnant unable to join them. How is this money assigned to a and B two wives more appropriate? How many dollars does A deserve? 90/(5+4) *5=$50 yuan? If you think so, you'll be fooled! The correct answer is 60 yuan. If you don't get it, think again. Here's a general question: Suppose Mrs. A worked for x days, Mrs. B worked for y days, and Mrs. C paid $90, how much does Mrs. a deserve? The input guarantees that two wives should be given non-negative integer dollars. Three wives work the same. Friendship Hint: There is a small trap in the subject. If the answer is wrong, please check the code carefully. Input

Enter the number of first behavior Data group T (T<=20). Only one row per group of data, containing three integers x, y, Z (1<=x, y<=10,1<=z<=1000).

Output

For each set of data, output an integer, that is, the amount that Mrs. a deserves (unit: yuan).

Sample Input

25 4 908) 4 123

Sample Output

60123

HINT

If using floating-point numbers, be careful about the error, and use rounding as much as possible when outputting.

Analysis: First of all, the first test of a mathematical study questions, in fact, the truth is very simple, analysis under the title of a deserved 60 yuan of the truth bar.

Originally was to three people to do together, now 2 people do, 2 people altogether spent 9 days, everyone work efficiency is the same, so the work is 9 days, so split up each person to do 3 days, but now C did not do, C of workload please a a share, share of how much money, a share 2 days, b Shared 1 days, so a was two-thirds, that is, 60 yuan;

Know the test instructions after it is very simple, a few lines of code can be done, need to pay a little attention to the amount of money will be multiplied by the proportion of the decimal, so to use a double type storage, the final rounding, with floor (apay+0.5) can be.

My code is as follows

#include <iostream>#include<cstdio>#include<cmath>using namespacestd;intMain () {intx, y, T, Z; CIN>>T;  while(t--) {cin>> x>>y>>Z; DoubleAve = (Double) (x + y)/3; DoubleAexp = X-ave;DoubleBexp = y-Ave; if(Aexp <0) Aexp =0; if(Bexp <0) Bexp =0; DoubleApay = (z* (aexp/(Aexp +( bexp))); cout<< Floor (apay+0.5) <<Endl; }    return 0;}

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1111: Three tips for people to solve problems