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.
- 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.
For each test case, output on a single line the minimum number of bags Chef needs.
Subtask 1 (100 points): Original constraints
Sample Input 1
3 2 3 5 10 1 2 3 5 3 3 4 4
Sample Output 1
1 2 3
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.
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)
October Long Challenge 2021
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- Digit Removal
- Yet another MEX problem
- Characteristic Polynomial Verification
- Chef at the Olympics
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- Three Boxes