# October Long Challenge Three Boxes

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## October Long Challenge Three Boxes

Chef has 33 boxes of sizes AA, BB, and CC respectively. He puts the boxes in bags of size DD (A≤B≤C≤DA≤B≤C≤D). Find the minimum number of bags Chef needs so that he can put each box in a bag. A bag can contain more than one box if the sum of sizes of boxes in the bag does not exceed the size of the bag.

### Input Format

• The first line contains TT denoting the number of test cases. Then the test cases follow.
• Each test case contains four integers AA, BB, CC, and DD on a single line denoting the sizes of the boxes and bags.

### Output Format

For each test case, output on a single line the minimum number of bags Chef needs.

### Constraints

• 1≤T≤1001≤T≤100
• 1≤A≤B≤C≤D≤1001≤A≤B≤C≤D≤100

Subtask 1 (100 points): Original constraints

```3
2 3 5 10
1 2 3 5
3 3 4 4
```

```1
2
3
```

### Explanation

Test case 11: The sum of sizes of boxes is 2+3+5=102+3+5=10 which is equal to the size of a bag. Hence Chef can put all three boxes in a single bag.

Test case 22: Chef can put boxes of size 11 and 33 in one bag and box of size 22 in another bag.

Test case 33: Chef puts all the boxes in separate bags as there is no way to put more than one box in a single bag.

CLICK BELOW!!

## Solution

Program Python:

```for _ in range (int(input())):
a,b,c,d=map(int,input().split())
if a+b+c<=d:
print(1)
elif a+c<=d:
print(2)
elif b+a<=d:
print(2)
else:
print(3)```