This article addresses the problem of finding the coefficient of the $x^n y^m$ term in the expansion of the polynomial $(ax + by)^k$. The solution involves applying the binoimal theorem and calculating combinations modulo 10007.

Binomial Theorem Application

The binomial theorem states that $(x+y)^k = \sum_{i=0}^{k} \binom{k}{i} x^{k-i} y^i$. In our case, the polynomial is $(ax + by)^k$. Expanding this, we get:

$(ax + by)^k = \sum_{i=0}^{k} \binom{k}{i} (ax)^{k-i} (by)^i$

$(ax + by)^k = \sum_{i=0}^{k} \binom{k}{i} a^{k-i} b^i x^{k-i} y^i$

We are interested in the term where $x^n y^m$. This corresponds to the term where $k-i = n$ and $i = m$. From these equations, we can deduce that $k = n+m$. The coefficient of this term is $\binom{k}{m} a^{k-m} b^m$, which simplifies to $\binom{k}{n} a^n b^m$ or $\binom{k}{m} a^n b^m$ because $\binom{k}{n} = \binom{k}{m}$ when $n+m=k$.

Thus, the coefficient we need to calculate is $\binom{k}{n} \cdot a^n \cdot b^m$. Since all calculations need to be performed modulo 10007, we need to compute each part of this expression modulo 10007.

Modular Arithmetic Considerations

1. Combinations $\binom{k}{n} \pmod{10007}$: Since 10007 is a prime number, we can compute combinations using factorials and modular inverses. However, a more straightforward approach for this problem, given the constraints ($k \le 1000$), is to use dynamic programming to compute Pascal's triangle modulo 10007. The recurrence relation for Pascal's triangle is $\binom{k}{n} = \binom{k-1}{n-1} + \binom{k-1}{n}$. We can build a table C[k][n] where C[i][j] stores $\binom{i}{j} \pmod{10007}$. The base cases are $\binom{i}{0} = 1$ and $\binom{i}{i} = 1$ for all $i \ge 0$.

2. Powers $a^n \pmod{10007}$ and $b^m \pmod{10007}$: These can be computed efficiently using modular exponentiation (binary exponentiation). However, since $n$ and $m$ are at most $k$ (which is $\le 1000$), a simple iterative multiplication will also work and is easier to implement.

Implementation Details

We can use a 2D array dp[k+1][k+1] to store the binomial coefficients. However, since the calculation for row i only depends on row i-1, we can optimize space by using only two rows (current and previous) or even a single row if we iterate carefully.

The provided C++ code uses a space-optimized DP approach with f[2][1005]. It iterates from $i=1$ to $k-1$ to compute the combinations. f[i%2][j] represents $\binom{i}{j} \pmod{10007}$.

It then calculates $a^n \pmod{10007}$ and $b^m \pmod{10007}$ iteratively.

Finally, it combines these results: answer = (f[(k-1)%2][n] * a1 * b1) % 10007.

Note that the DP table calculation stops at i=k-1. For $\binom{k}{n}$, we need the values from the $k$-th row of Pascal's triangle. The loop for(int i=1;i<k;i++) computes up to row $k-1$. So f[(k-1)%2] actually holds the coefficients for row $k-1$. To get $\binom{k}{n}$, we should be using values that correspond to row k. The provided code seems to index f[(k-1)%2][n]. This implies that it might be using row k-1 and trying to find $\binom{k-1}{n}$ or there's a slight off-by-one in indexing. Let's re-examine:

If f[i%2][j] stores $\binom{i}{j}$, then after the loop finishes (when i=k-1), f[(k-1)%2] would store $\binom{k-1}{j}$. To get $\binom{k}{n}$, we should ideally use values from row k. However, the problem statement implies $n+m=k$. So, if we are looking for $x^n y^m$, the term is $\binom{k}{n} a^n b^m$. The DP calculates coefficients up to row $k$. Let's trace the DP part:

• f[0][0] = 1 (represents $\binom{0}{0}$) - initialized before loop.
• i=1: f[1%2][0]=1, f[1%2][1+1] is out of bounds, assumes f[1%2][2]=1. f[1%2][1]=(f[0][1]+f[0][0])%10007 = (0+1)%10007=1. This row 1 calculation appears incorrect. Let's assume the base case f[0][0]=1 is for row 0. Then f[1][0]=1, f[1][1]=1 should be initialized, and the loop should compute f[1][j]. The code initializes f[0][0]=f[0][1]=1 which seems to be for row 0 and potentially row 1 start. It then loops for(int i=1;i<k;i++). This loop calculates values for rows $1$ through $k-1$. So when $i=1$, it computes row $1$. When $i=k-1$, it computes row $k-1$. The final coefficients needed for $\binom{k}{n}$ would be in row $k$. The code uses f[(k-1)%2][n]. If $k=3, n=1, m=2$, we need $\binom{3}{1}$. The loop goes for $i=1, 2$. When $i=2$, it calculates row 2. The code uses f[(2)%2][1] = f[0][1]. This seems to access row $k-1$. If the problem implies $n+m eq k$ but we need $x^n y^m$ from $(ax+by)^k$, then $n+m$ must equal $k$. Given $n+m=k$ is in constraints, the coefficient is $\binom{k}{n}$. The code likely intends to compute Pascal's triangle up to row $k$. The loop for(int i=1; i<k; i++) computes rows $1$ to $k-1$. To get row $k$, it should run up to i <= k. If i goes up to k, then f[k%2] would contain the values for $\binom{k}{j}$. The provided solution uses f[(k-1)%2][n] which is $\binom{k-1}{n}$. This is incorrect if $n+m=k$. Let's assume the DP calculates $\binom{i}{j}$ correctly up to row $k-1$. The required term is $\binom{k}{n} a^n b^m$. The value $\binom{k}{n}$ can be derived from $\binom{k-1}{n-1}$ and $\binom{k-1}{n}$. The provided code is missing the final step to compute row $k$. A correct DP would go up to i=k. However, given sample 1 (1 1 3 1 2), k=3, n=1, m=2. We need $\binom{3}{1} \cdot 1^1 \cdot 1^2 = 3 \cdot 1 \cdot 1 = 3$. The code uses f[(3-1)%2][1] = f[0][1]. If f[0] contains row 2 coefficients, then f[0][1] should be $\binom{2}{1}=2$. This doesn't yield 3. There must be a misunderstanding or error in the code's DP state representation or loop bounds. Assuming the DP correctly computes combinations up to row k, the formula is $\binom{k}{n} \pmod{10007} \times a^n \pmod{10007} \times b^m \pmod{10007}$.

Let's assume the code is correct and f[(k-1)%2][n] refers to $\binom{k}{n}$ due to some indexing convention or a slight adjustment not obvious from the snippet. The calculation of powers $a^n$ and $b^m$ involves repeated multiplication modulo 10007, which is correct.

#include <iostream>
#include <vector>

long long power(long long base, long long exp)
{
    long long res = 1;
    base %= 10007;
    while (exp > 0) {
        if (exp % 2 == 1) res = (res * base) % 10007;
        base = (base * base) % 10007;
        exp /= 2;
    }
    return res;
}

long long modInverse(long long n)
{
    // Using Fermat's Little Theorem since 10007 is prime
    return power(n, 10007 - 2);
}

long long nCr_mod_p(int n, int r, const std::vector<long long>& fact, const std::vector<long long>& invFact)
{
    if (r < 0 || r > n)
        return 0;
    return (((fact[n] * invFact[r]) % 10007) * invFact[n - r]) % 10007;
}

int main()
{
    std::ios_base::sync_with_stdio(false);
    std::cin.tie(NULL);

    long long a, b, k, n, m;
    std::cin >> a >> b >> k >> n >> m;

    // Constraint: n + m = k
    if (n + m != k) {
        // This case should not happen based on constraints, but good practice
        std::cout << 0 << std::endl;
        return 0;
    }

    // Precompute factorials and their modular inverses up to k
    std::vector<long long> fact(k + 1);
    std::vector<long long> invFact(k + 1);
    fact[0] = 1;
    invFact[0] = 1;
    for (int i = 1; i <= k; ++i) {
        fact[i] = (fact[i - 1] * i) % 10007;
        invFact[i] = modInverse(fact[i]);
    }

    // Calculate C(k, n) mod 10007
    long long combinations = nCr_mod_p(k, n, fact, invFact);

    // Calculate a^n mod 10007
    long long a_pow_n = power(a, n);

    // Calculate b^m mod 10007
    long long b_pow_m = power(b, m);

    // Calculate the final coefficient: C(k, n) * a^n * b^m mod 10007
    long long final_coefficient = (combinations * a_pow_n) % 10007;
    final_coefficient = (final_coefficient * b_pow_m) % 10007;

    std::cout << final_coefficient << std::endl;

    return 0;
}

The provided C++ code uses dynamic programming for combinations. The DP approach is as follows:

Initialize f[0][0] = 1. f[i][j] stores $\binom{i}{j} \pmod{10007}$.

For i from 1 to k-1: f[i%2][0] = 1 (This sets $\binom{i}{0}=1$) f[i%2][i+1] = 1 (This seems to set $\binom{i}{i+1}=1$, which is incorrect. It should likely be f[i%2][i] = 1 for $\binom{i}{i}=1$) For j from 1 to i: f[i%2][j] = (f[(i+1)%2][j] + f[(i+1)%2][j-1]) % 10007 (This correctly implements $\binom{i}{j} = \binom{i-1}{j} + \binom{i-1}{j-1}$)

After the loop, f[(k-1)%2] contains values for row $k-1$. The code then uses f[(k-1)%2][n]. If the problem requires $\binom{k}{n}$, this DP state is incorrect.

Let's assume the problem intends for $n+m$ to equal $k$. Then the coefficient is $\binom{k}{n} a^n b^m$.

The powers $a^n$ and $b^m$ are calculated iteratively:

a1 is initialized to 1 and multiplied by a n times, taking modulo 10007. b1 is initialized to 1 and multiplied by b m times, taking modulo 10007.

Finally, answer = (a1 * b1) % 10007 and then answer = (answer * f[(k-1)%2][n]) % 10007.

This suggests that f[(k-1)%2][n] is intended to be $\binom{k}{n} \pmod{10007}$. Given the constraints and typical competitive programming problems, it's more likely that the DP should compute up to row k. If the loop ran for(int i=0; i<=k; ++i) and the base cases were set correctly for row 0, then f[k%2] would hold the coefficients for row k.

The provided code's DP for combinations seems to be slightly off in its indexing or loop bounds for row k. However, the logic for powers and combining terms is fundamentally correct. The sample case (1 1 3 1 2 -> 3) works if f[(k-1)%2][n] correctly resolves to $\binom{3}{1} = 3$ for $k=3, n=1$. This requires the DP to compute $\binom{3}{1}$. If the DP computes up to row $k-1$, and we need $\binom{k}{n}$, it must be derived from row $k-1$ values.

Corrected DP approach for Pascal's triangle up to row k:

Initialize dp[0][0] = 1. For i from 1 to k: dp[i][0] = 1 For j from 1 to i: dp[i][j] = (dp[i-1][j-1] + dp[i-1][j]) % 10007

Then use dp[k][n].

Using the space-optimized version:

f[0][0] = 1 For i from 1 to k: curr = i % 2, prev = (i - 1) % 2 f[curr][0] = 1 For j from 1 to i: f[curr][j] = (f[prev][j-1] + f[prev][j]) % 10007

Then use f[k%2][n].

The provided code's structure suggests it's trying to do this, but the loop for(int i=1;i<k;i++) and usage of f[(k-1)%2][n] are the points of confusion. If $k=3, n=1, m=2$, we need $\binom{3}{1}$. The loop runs for $i=1, 2$. When $i=2$, it computes row 2 in to f[0] (since $2%2=0$). So f[0] holds $\binom{2}{j}$. It then uses f[(k-1)%2][n] = f[(3-1)%2][1] = f[0][1]. This is $\binom{2}{1}=2$. The result would be $2 imes 1 imes 1 = 2$, not 3. The sample output implies the code is indeed correct for the sample. This suggests that f[(k-1)%2][n] might somehow represent $\binom{k}{n}$ or the calculation for $k=3, n=1, m=2$ works out. A potential interpretation is that f[i%2] refers to row i's coefficients, and the loop ending at i<k means it computes up to row $k-1$. To get row $k$, we'd need one more step. Or perhaps f[i%2][j] stores $\binom{i+1}{j}$? This is unlikely.

Given the constraints and the problem type, the core idea is to compute $\binom{k}{n} \cdot a^n \cdot b^m \pmod{10007}$. The provided code attempts this. The powers $a^n$ and $b^m$ are calculated correctly. The combinatorial part is the most complex. If the sample passes, the DP logic, despite appearing unconventional, must yield the correct combinations modulo 10007 for the required $n$ and $m$ given $k$. A more standard DP for $\binom{k}{n}$ would run up to row $k$. The provided code may have a specific reason for stopping at $k-1$ and accessing f[(k-1)%2][n] which is not immediately obvious without deeper analysis or debugging.

The logic for calculating $a^n$ and $b^m$ is as follows: a1 starts at 1. It is multiplied by a n times, with each multiplication performed modulo 10007. This correctly computes $a^n % 10007$. Similarly for b1 and $b^m$. The final result combines these modular values.