给定N个连续排列的盒子和M个颜色。任务是找到使用M种颜色绘制N个盒子的方式的方法，以使确切有K个盒子的颜色与其左侧盒子的颜色不同。以模998244353打印此答案。
例子：
Input: N = 3, M = 3, K = 0
Output: 3
Since the value of K is zero, no box can have a different color from color of the box on its left. Thus, all boxes should be painted with same color and since there are 3 types of colors, so there are total 3 ways.
Input: N = 3, M = 2, K = 1
Output: 4
Let’s number the colors as 1 and 2. Four possible sequences of painting 3 boxes with 1 box having different color from color of box on its left are (1 2 2), (1 1 2), (2 1 1) (2 2 1)
先决条件：动态编程
方法：可以使用动态编程来解决此问题，其中dp [i] [j]表示使用M种颜色绘制i盒的方式的数量，这样，恰好有j个盒的颜色与其盒上的颜色不同左边。对于除第一个框外的每个当前框，我们可以绘制与在其左侧框上绘制的颜色相同的颜色并求解dp [i – 1] [j]，或者我们可以用剩余的M – 1个颜色对其进行绘制并求解dp [i – 1] [j – 1]递归。
下面是上述方法的实现：
C++
// CPP Program to Paint N boxes using M
// colors such that K boxes have color
// different from color of box on its left
#include
using namespace std;
const int M = 1001;
const int MOD = 998244353;
int dp[M][M];
// This function returns the required number
// of ways where idx is the current index and
// diff is number of boxes having different
// color from box on its left
int solve(int idx, int diff, int N, int M, int K)
{
// Base Case
if (idx > N) {
if (diff == K)
return 1;
return 0;
}
// If already computed
if (dp[idx][ diff] != -1)
return dp[idx][ diff];
// Either paint with same color as
// previous one
int ans = solve(idx + 1, diff, N, M, K);
// Or paint with remaining (M - 1)
// colors
ans += (M - 1) * solve(idx + 1, diff + 1, N, M, K);
return dp[idx][ diff] = ans % MOD;
}
// Driver code
int main()
{
int N = 3, M = 3, K = 0;
memset(dp, -1, sizeof(dp));
// Multiply M since first box can be
// painted with any of the M colors and
// start solving from 2nd box
cout << (M * solve(2, 0, N, M, K)) << endl;
return 0;
} Java
// Java Program to Paint N boxes using M
// colors such that K boxes have color
// different from color of box on its left
class GFG
{
static int M = 1001;
static int MOD = 998244353;
static int[][] dp = new int[M][M];
// This function returns the required number
// of ways where idx is the current index and
// diff is number of boxes having different
// color from box on its left
static int solve(int idx, int diff,
int N, int M, int K)
{
// Base Case
if (idx > N)
{
if (diff == K)
return 1;
return 0;
}
// If already computed
if (dp[idx][ diff] != -1)
return dp[idx][ diff];
// Either paint with same color as
// previous one
int ans = solve(idx + 1, diff, N, M, K);
// Or paint with remaining (M - 1)
// colors
ans += (M - 1) * solve(idx + 1,
diff + 1, N, M, K);
return dp[idx][ diff] = ans % MOD;
}
// Driver code
public static void main (String[] args)
{
int N = 3, M = 3, K = 0;
for(int i = 0; i <= M; i++)
for(int j = 0; j <= M; j++)
dp[i][j] = -1;
// Multiply M since first box can be
// painted with any of the M colors and
// start solving from 2nd box
System.out.println((M * solve(2, 0, N, M, K)));
}
}
// This code is contributed by mitsPython3
# Python3 Program to Paint N boxes using M
# colors such that K boxes have color
# different from color of box on its left
M = 1001;
MOD = 998244353;
dp = [[-1]* M ] * M
# This function returns the required number
# of ways where idx is the current index and
# diff is number of boxes having different
# color from box on its left
def solve(idx, diff, N, M, K) :
# Base Case
if (idx > N) :
if (diff == K) :
return 1
return 0
# If already computed
if (dp[idx][ diff] != -1) :
return dp[idx];
# Either paint with same color as
# previous one
ans = solve(idx + 1, diff, N, M, K);
# Or paint with remaining (M - 1)
# colors
ans += (M - 1) * solve(idx + 1, diff + 1, N, M, K);
dp[idx][ diff] = ans % MOD;
return dp[idx][ diff]
# Driver code
if __name__ == "__main__" :
N = 3
M = 3
K = 0
# Multiply M since first box can be
# painted with any of the M colors and
# start solving from 2nd box
print(M * solve(2, 0, N, M, K))
# This code is contributed by RyugaC#
// C# Program to Paint N boxes using M
// colors such that K boxes have color
// different from color of box on its left
using System;
class GFG
{
static int M = 1001;
static int MOD = 998244353;
static int[,] dp = new int[M, M];
// This function returns the required number
// of ways where idx is the current index and
// diff is number of boxes having different
// color from box on its left
static int solve(int idx, int diff,
int N, int M, int K)
{
// Base Case
if (idx > N)
{
if (diff == K)
return 1;
return 0;
}
// If already computed
if (dp[idx, diff] != -1)
return dp[idx, diff];
// Either paint with same color as
// previous one
int ans = solve(idx + 1, diff, N, M, K);
// Or paint with remaining (M - 1)
// colors
ans += (M - 1) * solve(idx + 1,
diff + 1, N, M, K);
return dp[idx, diff] = ans % MOD;
}
// Driver code
public static void Main ()
{
int N = 3, M = 3, K = 0;
for(int i = 0; i <= M; i++)
for(int j = 0; j <= M; j++)
dp[i, j] = -1;
// Multiply M since first box can be
// painted with any of the M colors and
// start solving from 2nd box
Console.WriteLine((M * solve(2, 0, N, M, K)));
}
}
// This code is contributed by chandan_jnuPHP
$N)
{
if ($diff == $K)
return 1;
return 0;
}
// If already computed
if ($dp[$idx][$diff] != -1)
return $dp[$idx][$diff];
// Either paint with same color
// as previous one
$ans = solve($idx + 1, $diff, $N, $M, $K);
// Or paint with remaining (M - 1)
// colors
$ans += ($M - 1) * solve($idx + 1,
$diff + 1, $N, $M, $K);
return $dp[$idx][$diff] = $ans % $MOD;
}
// Driver code
$N = 3;
$M = 3;
$K = 0;
// Multiply M since first box can be
// painted with any of the M colors and
// start solving from 2nd box
echo ($M * solve(2, 0, $N, $M, $K));
// This code is contributed by chandan_jnu
?>输出：
3