Competitive Programming #2: [Count numbers which can be constructed using two numbers]

Here is the problem:

Given three positive integers x, y and n, the task is to find count of all numbers from 1 to n that can be formed using x and y. A number can be formed using x and y if we can get it by adding any number of occurrences of x and/or y.

Input:
The first line of input contains an integer T denoting the number of test cases. Then T test cases follow. The first line of each test case contains the integers x, y and n respectively.

Output:
Print the count of all numbers between 1 and n which can be formed using x and y. Print the answer for each test case in a new line.

Constraints:
1<= T <=100
1<= x, y, n <=100000

Example:
Input:
2
2 3 10
5 7 10
Output:
9
3
Explanation:
Input : n = 10, x = 2, y = 3
Output : 9
We can form 9 out of 10 numbers using 2 and 3 and 10 2 = 2, 3 = 3, 4 = 2+2, 5 = 2+3, 6 = 3+3 7 = 2+2+3, 8 = 3+3+2, 9 = 3+3+3 , 10 = 3+3+2+2.

This problem is more mathematical. Here is explanation:
 My Code:
In this code , the most important thing is the vector . We remember the values in the vector that are derived from x and y already
 
Let's see the example below: (Follow the code where start from 'for')

x=1; y= 2; n=5;
>> i=min(x,y) = 1;  IS i<=5 ? Yes #continue;
>> IS i>=1 or i>=2 ? Yes #continue;
>> IS i>=1 and Numbers[i-1]=true ? NO
>> IS i>=2 and Numbers[i-2]=true ? NO
>> IS i=1 ? YES #counter+1; #Number[i]=true;

Continue like this to i=5;