Three Boxes solution - Codechef

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Chef has 3$3$ boxes of sizes A$A$, B$B$, and C$C$ respectively. He puts the boxes in bags of size D$D$ (A≤B≤C≤D$A\le B\le C\le D$). 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 T$T$ denoting the number of test cases. Then the test cases follow.
• Each test case contains four integers A$A$, B$B$, C$C$, and D$D$ 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≤100$1\le T\le 100$
• 1≤A≤B≤C≤D≤100$1\le A\le B\le C\le D\le 100$

Subtasks

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

Explanation

Test case 1$1$: The sum of sizes of boxes is 2+3+5=10$2+3+5=10$ which is equal to the size of a bag. Hence Chef can put all three boxes in a single bag.

Test case 2$2$: Chef can put boxes of size 1$1$ and 3$3$ in one bag and box of size 2$2$ in another bag.

Test case 3$3$: Chef puts all the boxes in separate bags as there is no way to put more than one box in a single bag.

Three Boxes solution :

#include <iostream>

using namespace std;

int main() {

// your code goes here

int t;

cin>>t;

while(t--){

    int a, b, c, d;

    cin>>a>>b>>c>>d;

    if((a+b+c)<=d){

        cout<<1<<endl;

    }

    else if((a+b)<=d){

        cout<<2<<endl;

    }

    else{

        cout<<3<<endl;

    }

}

return 0;

}