# Hackerrank Mathematics- Die Hard 3 Solution

** Simon**: On the fountain, there should be 2 jugs - a 5 gallon and a 3 gallon. Do you see them? Fill one of the jugs with exactly 4 gallons of water and place it on the scale and the timer will stop. You must be precise; one ounce more or less will result in detonation. If you’re still alive in 5 minutes, we’ll speak.

** Bruce**: Wait, wait a second. I don’t get it. Do you get it?

** Samuel**: No.

** Bruce**: Get the jugs. Obviously, we can’t ﬁll the 3 gallon jug with 4 gallons of water.

** Samuel**: Obviously.

** Bruce**: All right. I know, here we go. We ﬁll the 3 gallon jug exactly to the top, right?

** Samuel**: Uh huh.

** Bruce**: Okay, now we pour this 3 gallons into the 5 gallon jug, giving us exactly 3 gallons in the 5 gallon jug, right?

** Samuel**: Right, then what?

** Bruce**: We take the 3 gallon jug and ﬁll it a third of the way...

** Samuel**: No! He said, “Be precise.” Exactly 4 gallons.

** Bruce**: Damn! Every cop within 50 miles is running his ass off and I’m out here playing a kids' games in the park.

** Samuel**: Hey, you want to focus on the problem at hand?

Given 2 jugs of capacity *a* and *b* gallons, and an infinite supply of water, can you fill one of the jugs with exactly *c* gallons of water ?

**Input Format**

First line contains the number of testcases *T*. *T* lines follow.

Each line contains 3 space separated integers *a*, *b* and *c* . *a* and *b* indicate the capacity of the two jugs respectively, and *c* denotes the exact capacity with which one of the jugs should be filled.

**Output Format**

For each test case, print "YES" (in a new line) if one of the jugs can be filled with exactly *c* gallons of water and "NO" (in a new line) if they cannot be filled. ( quotes are for clarity )

**Constraints**

1 ≤ a, b, c ≤ 10^{3}

1 ≤ T ≤ 100

**Sample Input**

```
2
5 3 4
3 6 4
```

**Sample Output**

```
YES
NO
```

**Explanation**

Bruce can do the following, fill jug *a* with 5 gallons.

```
a = 5, b = 0
```

Now, he can fill jug *b* with 3 gallons from jug *a*.

```
a = 2, b = 3
```

He can empty jug *b* and empty 2 gallons from jug *a* to jug *b*.

```
a = 0, b = 2
```

Now, he can fill jug *a* with 5 gallons and fill jug *b* with 1 gallon from jug *a*. This results in jug *a* containing exactly 4 gallons of water.

```
a = 5, b = 2
a = 4, b = 3
```

In the second testcase, it is impossible to generate 4 gallons of water, hence `NO`

.

### Solution in Python

```
def solve(a,b,c):
A= a
x= 0
while True:
x = a-(b-x)
if x==A:
return "NO"
if x==c:
return "YES"
while(x>b):
x = x-b
if x==c:
return "YES"
for i in range(int(input())):
a, b , c= map(int, input("").split())
x = [a,b]
x.sort()
b,a = x
print(solve(a,b,c))
```