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# C++ program to count in how many ways we can paint blocks with two conditions

Suppose we have three numbers N, M and K. Consider there are N blocks they are arranged in a row. We consider following two ways to paint them. The paints of two blocks different if and only if the blocks are painted in different colors in following two ways −

For each block, use one of the M colors to paint it. (it is not mandatory to use all colors)

There may be at most K pairs of adjacent blocks that are painted in the same color.

If the answer is too large, return result mod 998244353.

So, if the input is like N = 3; M = 2; K = 1, then the output will be 6, because we can paint them in these different formats 112, 121, 122, 211, 212, and 221.

## Steps

To solve this, we will follow these steps −

maxm := 2^6 + 5 p := 998244353 Define two large arrays fac and inv or size maxm Define a function ppow(), this will take a, b, p, ans := 1 mod p a := a mod p while b is non-zero, do: if b is odd, then: ans := ans * a mod p a := a * a mod p b := b/2 return ans Define a function C(), this will take n, m, if m < 0 or m > n, then: return 0 return fac[n] * inv[m] mod p * inv[n - m] mod p From the main method, do the following fac[0] := 1 for initialize i := 1, when i < maxm, update (increase i by 1), do: fac[i] := fac[i - 1] * i mod p inv[maxm - 1] := ppow(fac[maxm - 1], p - 2, p) for initialize i := maxm - 2, when i >= 0, update (decrease i by 1), do: inv[i] := (i + 1) * inv[i + 1] mod p ans := 0 for initialize i := 0, when i <= k, update (increase i by 1), do: t := C(n - 1, i) tt := m * ppow(m - 1, n - i - 1, p) ans := (ans + t * tt mod p) mod p return ans

## Example

Let us see the following implementation to get better understanding −

#include <bits/stdc++.h> using namespace std; const long maxm = 2e6 + 5; const long p = 998244353; long fac[maxm], inv[maxm]; long ppow(long a, long b, long p){ long ans = 1 % p; a %= p; while (b){ if (b & 1) ans = ans * a % p; a = a * a % p; b >>= 1; } return ans; } long C(long n, long m){ if (m < 0 || m > n) return 0; return fac[n] * inv[m] % p * inv[n - m] % p; } long solve(long n, long m, long k){ fac[0] = 1; for (long i = 1; i < maxm; i++) fac[i] = fac[i - 1] * i % p; inv[maxm - 1] = ppow(fac[maxm - 1], p - 2, p); for (long i = maxm - 2; i >= 0; i--) inv[i] = (i + 1) * inv[i + 1] % p; long ans = 0; for (long i = 0; i <= k; i++){ long t = C(n - 1, i); long tt = m * ppow(m - 1, n - i - 1, p) % p; ans = (ans + t * tt % p) % p; } return ans; } int main(){ int N = 3; int M = 2; int K = 1; cout << solve(N, M, K) << endl; }

## Input

3, 2, 1

## Output

6

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