C++ program to count the number of coloring schemes that satisfy two conditions

Suppose we have three numbers N, M and K. Consider there are N blocks, arranged in a row. We consider the following two ways of coloring. Two blocks are colored differently if and only if the blocks in the following two ways are colored in different colors: -

• For each block, use one of M colors to color (not necessarily using all colors)

• There may be at most K pairs of adjacent blocks painted with the same color

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

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

Steps

To solve this problem, we will follow the following 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 implementation below for 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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