# 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

Updated on: 03-Mar-2022

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