PROBLEM LINK:
Author: Abhishek Ghosh
Tester: Rahul Sahay
Editorialist: Gaurav Kumar, Md. Sami Khan
DIFFICULTY
Easy
PREREQUISITES
Math, Distribution, DP, Precomputation
PROBLEM
You have been given a weight of N kilos. The measure will be accurate only if 3 calibration weights are used. You need to put 3 weights such that the sum equals N kilos, given that the calibration weights are in multiples of K.
QUICK EXPLANATION

We have to divide N into 3 multiples of k (k{a}, k{b}, k{c}) such that:
k{a} + k{b} + k{c} = N \implies a+b+c = \frac{N}{k}

The problem can be related to a common problem, distributing \frac{N}{k} identical balls into 3 identical boxes such that no box should be empty.

Let f(N, m) be a function to distribute N identical balls into m identical boxes. In this particular problem:
f(N, 3)= f(N3,1) + f(N3, 2) + f(N3, 3)
This recurrence relation is used to solve the problem.
EXPLANATION
In the following problem, we have to split N into 3 multiples of k. None of the multiples should be 0.
If N is not divisible by K then the answer will simply be 0. Otherwise, we need to distribute \frac{N}{k} identical balls among 3 identical boxes, such that each should get at least one.
f(N, m) is a function to divide N into m multiples or to put N balls into m boxes such that no box is empty.
We give 1 ball to each of the 3 identical boxes and now we are left with N3 balls to distribute without any restriction.
Now, we can either give all the N3 balls to 1 box, or we can give the N3 balls to 2 boxes, or we can give N3 balls to all the 3 boxes which is nothing but f(N3, 3), and similar to our initial condition that we started with.
Thus, f(N, 3) has 3 cases:
f(N, 3)= f(N3,1)+f(N3, 2)+f(N3, 3)
Case 1
N3 is given to just 1 box.
This can be done only in 1 way.
\implies f(N3,1) = 1 for all N >= 3
Case 2
N3 is given to just 2 boxes.
For N=3:
f(0,2)=0For N=4:
f(1,2)=0For N=5:
f(2,2)=1 {[1,1]}For N=6:
f(3,2)=1 {[2,1]}For N=7:
f(4,2)=2 {[3,1],[2,2]}For N=8:
f(5,2)=2 {[4,1], [3,2]}For N=9:
f(6,2)=3 {[5,1], [4,2], [3,3]}For N=10:
f(7,2)=3 {[6,1], [5,2], [4,3]}By recognizing the simple logic, or understanding the pattern we can say that this can be done in \frac{N3}{2} ways.
\implies f(N3,2)=\frac{N3}{2} \; \forall N \geq 3
Case 3 (Actual case)
N3 is given to all 3 multiples.
This can further be broken into 3 cases.
f(N3, 3) = f(N6,1) + f(N6, 2) + f(N6, 3)
This is the initial case that we started with.f(N, 3)= 1 + \frac{N3}{2} + f(N3, 3)
Approach 1:
We can call the recurrence relation for each value of N. But in this approach we will have to calculate the value of f(Nr, 3) multiple times, where 0<r<N. This approach is acceptable for smaller values of N and K.
Time complexity:
O(T\mathord{\cdot}N) where T is the number of test cases.
Approach 2:
We can use precomputation and store the final answer for each N in an array; the i^{th} index of the array gives the answer for N=i
Time complexity:
O(T + N) where T is the number of test cases.
SOLUTIONS
Setter's Solution
#include <bits/stdc++.h>
#define mod 1000000007
using namespace std;
const int maxN = 1e6;
int dp[maxN+1];
int32_t main() {
#ifndef ONLINE_JUDGE
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
// precompute till 10^6
dp[0] = dp[1] = dp[2] = 0;
for(int i = 3; i <= maxN; ++i) {
dp[i] = (1 + (i3)/2 + dp[i3]) % mod;
}
int T = 1;
cin >> T;
while(T) {
int n, k;
cin >> n >> k;
if(n%k) {
cout << 0 << "\n";
continue;
}
n = n/k;
cout << dp[n] << "\n";
}
return 0;
}