有 P 型的“p”球、Q 型的“q”球和 R 型的“r”球。使用这些球,我们要创建一条直线,使得没有两个相同类型的球相邻。
例子 :
Input : p = 1, q = 1, r = 0
Output : 2
There are only two arrangements PQ and QP
Input : p = 1, q = 1, r = 1
Output : 6
There are only six arrangements PQR, QPR,
QRP, RQP, PRQ and RPQ
Input : p = 2, q = 1, r = 1
Output : 6
There are only six arrangements PQRP, QPRP,
PRQP, RPQP, PRPQ and PQPR
我们强烈建议您在继续解决方案之前单击此处进行练习。
天真的解决方案:
这个问题的天真解决方案是递归解决方案。我们递归调用三种情况
1) 最后放置的球是 P 型球
2) 最后放置的球是 Q 型球
3) 最后放置的球是 R 型
下面是上述想法的实现。
C++
// C++ program to count number of ways to arrange three
// types of balls such that no two balls of same color
// are adjacent to each other
#include
using namespace std;
// Returns count of arrangements where last placed ball is
// 'last'. 'last' is 0 for 'p', 1 for 'q' and 2 for 'r'
int countWays(int p, int q, int r, int last)
{
// if number of balls of any color becomes less
// than 0 the number of ways arrangements is 0.
if (p<0 || q<0 || r<0)
return 0;
// If last ball required is of type P and the number
// of balls of P type is 1 while number of balls of
// other color is 0 the number of ways is 1.
if (p==1 && q==0 && r==0 && last==0)
return 1;
// Same case as above for 'q' and 'r'
if (p==0 && q==1 && r==0 && last==1)
return 1;
if (p==0 && q==0 && r==1 && last==2)
return 1;
// if last ball required is P and the number of ways is
// the sum of number of ways to form sequence with 'p-1' P
// balls, q Q Balls and r R balls ending with Q and R.
if (last==0)
return countWays(p-1,q,r,1) + countWays(p-1,q,r,2);
// Same as above case for 'q' and 'r'
if (last==1)
return countWays(p,q-1,r,0) + countWays(p,q-1,r,2);
if (last==2)
return countWays(p,q,r-1,0) + countWays(p,q,r-1,1);
}
// Returns count of required arrangements
int countUtil(int p, int q, int r)
{
// Three cases arise:
return countWays(p, q, r, 0) + // Last required balls is type P
countWays(p, q, r, 1) + // Last required balls is type Q
countWays(p, q, r, 2); // Last required balls is type R
}
// Driver code to test above
int main()
{
int p = 1, q = 1, r = 1;
printf("%d", countUtil(p, q, r));
return 0;
}
Java
// Java program to count number
// of ways to arrange three types of
// balls such that no two balls of
// same color are adjacent to each other
class GFG {
// Returns count of arrangements
// where last placed ball is
// 'last'. 'last' is 0 for 'p',
// 1 for 'q' and 2 for 'r'
static int countWays(int p, int q, int r, int last)
{
// if number of balls of any
// color becomes less than 0
// the number of ways arrangements is 0.
if (p < 0 || q < 0 || r < 0)
return 0;
// If last ball required is
// of type P and the number
// of balls of P type is 1
// while number of balls of
// other color is 0 the number
// of ways is 1.
if (p == 1 && q == 0 && r == 0 && last == 0)
return 1;
// Same case as above for 'q' and 'r'
if (p == 0 && q == 1 && r == 0 && last == 1)
return 1;
if (p == 0 && q == 0 && r == 1 && last == 2)
return 1;
// if last ball required is P
// and the number of ways is
// the sum of number of ways
// to form sequence with 'p-1' P
// balls, q Q Balls and r R balls
// ending with Q and R.
if (last == 0)
return countWays(p - 1, q, r, 1) +
countWays(p - 1, q, r, 2);
// Same as above case for 'q' and 'r'
if (last == 1)
return countWays(p, q - 1, r, 0) +
countWays(p, q - 1, r, 2);
if (last == 2)
return countWays(p, q, r - 1, 0) +
countWays(p, q, r - 1, 1);
return 0;
}
// Returns count of required arrangements
static int countUtil(int p, int q, int r) {
// Three cases arise:
return countWays(p, q, r, 0) + // Last required balls is type P
countWays(p, q, r, 1) + // Last required balls is type Q
countWays(p, q, r, 2); // Last required balls is type R
}
// Driver code
public static void main(String[] args)
{
int p = 1, q = 1, r = 1;
System.out.print(countUtil(p, q, r));
}
}
// This code is contributed by Anant Agarwal.
Python3
# Python3 program to count
# number of ways to arrange
# three types of balls such
# that no two balls of same
# color are adjacent to each
# other
# Returns count of arrangements
# where last placed ball is
# 'last'. 'last' is 0 for 'p',
# 1 for 'q' and 2 for 'r'
def countWays(p, q, r, last):
# if number of balls of
# any color becomes less
# than 0 the number of
# ways arrangements is 0.
if (p < 0 or q < 0 or r < 0):
return 0;
# If last ball required is
# of type P and the number
# of balls of P type is 1
# while number of balls of
# other color is 0 the number
# of ways is 1.
if (p == 1 and q == 0 and
r == 0 and last == 0):
return 1;
# Same case as above
# for 'q' and 'r'
if (p == 0 and q == 1 and
r == 0 and last == 1):
return 1;
if (p == 0 and q == 0 and
r == 1 and last == 2):
return 1;
# if last ball required is P
# and the number of ways is
# the sum of number of ways
# to form sequence with 'p-1' P
# balls, q Q Balls and r R
# balls ending with Q and R.
if (last == 0):
return (countWays(p - 1, q, r, 1) +
countWays(p - 1, q, r, 2));
# Same as above case
# for 'q' and 'r'
if (last == 1):
return (countWays(p, q - 1, r, 0) +
countWays(p, q - 1, r, 2));
if (last == 2):
return (countWays(p, q, r - 1, 0) +
countWays(p, q, r - 1, 1));
# Returns count of
# required arrangements
def countUtil(p, q, r):
# Three cases arise:
# Last required balls is type P
# Last required balls is type Q
# Last required balls is type R
return (countWays(p, q, r, 0) +
countWays(p, q, r, 1) +
countWays(p, q, r, 2));
# Driver Code
p = 1;
q = 1;
r = 1;
print(countUtil(p, q, r));
# This code is contributed by mits
C#
// C# program to count number
// of ways to arrange three types of
// balls such that no two balls of
// same color are adjacent to each other
using System;
class GFG {
// Returns count of arrangements
// where last placed ball is
// 'last'. 'last' is 0 for 'p',
// 1 for 'q' and 2 for 'r'
static int countWays(int p, int q,
int r, int last)
{
// if number of balls of any
// color becomes less than 0
// the number of ways
// arrangements is 0.
if (p < 0 || q < 0 || r < 0)
return 0;
// If last ball required is
// of type P and the number
// of balls of P type is 1
// while number of balls of
// other color is 0 the number
// of ways is 1.
if (p == 1 && q == 0 && r == 0
&& last == 0)
return 1;
// Same case as above for 'q' and 'r'
if (p == 0 && q == 1 && r == 0
&& last == 1)
return 1;
if (p == 0 && q == 0 && r == 1
&& last == 2)
return 1;
// if last ball required is P
// and the number of ways is
// the sum of number of ways
// to form sequence with 'p-1' P
// balls, q Q Balls and r R balls
// ending with Q and R.
if (last == 0)
return countWays(p - 1, q, r, 1) +
countWays(p - 1, q, r, 2);
// Same as above case for 'q' and 'r'
if (last == 1)
return countWays(p, q - 1, r, 0) +
countWays(p, q - 1, r, 2);
if (last == 2)
return countWays(p, q, r - 1, 0) +
countWays(p, q, r - 1, 1);
return 0;
}
// Returns count of required arrangements
static int countUtil(int p, int q, int r)
{
// Three cases arise:
// 1. Last required balls is type P
// 2. Last required balls is type Q
// 3. Last required balls is type R
return countWays(p, q, r, 0) +
countWays(p, q, r, 1) +
countWays(p, q, r, 2);
}
// Driver code
public static void Main()
{
int p = 1, q = 1, r = 1;
Console.Write(countUtil(p, q, r));
}
}
// This code is contributed by nitin mittal.
PHP
Javascript
C++
// C++ program to count number of ways to arrange three
// types of balls such that no two balls of same color
// are adjacent to each other
#include
using namespace std;
#define MAX 100
// table to store to store results of subproblems
int dp[MAX][MAX][MAX][3];
// Returns count of arrangements where last placed ball is
// 'last'. 'last' is 0 for 'p', 1 for 'q' and 2 for 'r'
int countWays(int p, int q, int r, int last)
{
// if number of balls of any color becomes less
// than 0 the number of ways arrangements is 0.
if (p<0 || q<0 || r<0)
return 0;
// If last ball required is of type P and the number
// of balls of P type is 1 while number of balls of
// other color is 0 the number of ways is 1.
if (p==1 && q==0 && r==0 && last==0)
return 1;
// Same case as above for 'q' and 'r'
if (p==0 && q==1 && r==0 && last==1)
return 1;
if (p==0 && q==0 && r==1 && last==2)
return 1;
// If this subproblem is already evaluated
if (dp[p][q][r][last] != -1)
return dp[p][q][r][last];
// if last ball required is P and the number of ways is
// the sum of number of ways to form sequence with 'p-1' P
// balls, q Q Balls and r R balls ending with Q and R.
if (last==0)
dp[p][q][r][last] = countWays(p-1,q,r,1) + countWays(p-1,q,r,2);
// Same as above case for 'q' and 'r'
else if (last==1)
dp[p][q][r][last] = countWays(p,q-1,r,0) + countWays(p,q-1,r,2);
else //(last==2)
dp[p][q][r][last] = countWays(p,q,r-1,0) + countWays(p,q,r-1,1);
return dp[p][q][r][last];
}
// Returns count of required arrangements
int countUtil(int p, int q, int r)
{
// Initialize 'dp' array
memset(dp, -1, sizeof(dp));
// Three cases arise:
return countWays(p, q, r, 0) + // Last required balls is type P
countWays(p, q, r, 1) + // Last required balls is type Q
countWays(p, q, r, 2); // Last required balls is type R
}
// Driver code to test above
int main()
{
int p = 1, q = 1, r = 1;
printf("%d", countUtil(p, q, r));
return 0;
}
Java
// Java program to count number
// of ways to arrange three
// types of balls such that no
// two balls of same color
// are adjacent to each other
import java.util.Arrays;
class GFG
{
static final int MAX = 100;
// table to store to store results of subproblems
static int dp[][][][] = new int[MAX][MAX][MAX][3];
// Returns count of arrangements
// where last placed ball is
// 'last'. 'last' is 0 for 'p',
// 1 for 'q' and 2 for 'r'
static int countWays(int p, int q, int r, int last)
{
// if number of balls of any
// color becomes less than 0
// the number of ways arrangements is 0.
if (p < 0 || q < 0 || r < 0)
return 0;
// If last ball required is
// of type P and the number
// of balls of P type is 1
// while number of balls of
// other color is 0 the number
// of ways is 1.
if (p == 1 && q == 0 && r == 0 && last == 0)
return 1;
// Same case as above for 'q' and 'r'
if (p == 0 && q == 1 && r == 0 && last == 1)
return 1;
if (p == 0 && q == 0 && r == 1 && last == 2)
return 1;
// If this subproblem is already evaluated
if (dp[p][q][r][last] != -1)
return dp[p][q][r][last];
// if last ball required is P and
// the number of ways is the sum
// of number of ways to form sequence
// with 'p-1' P balls, q Q balss and
// r R balls ending with Q and R.
if (last == 0)
dp[p][q][r][last] = countWays(p - 1, q, r, 1) +
countWays(p - 1, q, r, 2);
// Same as above case for 'q' and 'r'
else if (last == 1)
dp[p][q][r][last] = countWays(p, q - 1, r, 0) +
countWays(p, q - 1, r, 2);
//(last==2)
else
dp[p][q][r][last] = countWays(p, q, r - 1, 0) +
countWays(p, q, r - 1, 1);
return dp[p][q][r][last];
}
// Returns count of required arrangements
static int countUtil(int p, int q, int r)
{
// Initialize 'dp' array
for (int[][][] row : dp)
{
for (int[][] innerRow : row)
{
for (int[] innerInnerRow : innerRow)
{
Arrays.fill(innerInnerRow, -1);
}
}
};
// Three cases arise:
return countWays(p, q, r, 0) + // Last required balls is type P
countWays(p, q, r, 1) + // Last required balls is type Q
countWays(p, q, r, 2); // Last required balls is type R
}
// Driver code
public static void main(String[] args)
{
int p = 1, q = 1, r = 1;
System.out.print(countUtil(p, q, r));
}
}
// This code is contributed by Anant Agarwal.
C#
// C# program to count number
// of ways to arrange three
// types of balls such that no
// two balls of same color
// are adjacent to each other
using System;
class GFG
{
static int MAX = 101;
// table to store to store results of subproblems
static int[,,,] dp = new int[MAX, MAX, MAX, 4];
// Returns count of arrangements
// where last placed ball is
// 'last'. 'last' is 0 for 'p',
// 1 for 'q' and 2 for 'r'
static int countWays(int p, int q, int r, int last)
{
// if number of balls of any
// color becomes less than 0
// the number of ways arrangements is 0.
if (p < 0 || q < 0 || r < 0)
return 0;
// If last ball required is
// of type P and the number
// of balls of P type is 1
// while number of balls of
// other color is 0 the number
// of ways is 1.
if (p == 1 && q == 0 && r == 0 && last == 0)
return 1;
// Same case as above for 'q' and 'r'
if (p == 0 && q == 1 && r == 0 && last == 1)
return 1;
if (p == 0 && q == 0 && r == 1 && last == 2)
return 1;
// If this subproblem is already evaluated
if (dp[p, q, r, last] != -1)
return dp[p, q, r, last];
// if last ball required is P and
// the number of ways is the sum
// of number of ways to form sequence
// with 'p-1' P balls, q Q balss and
// r R balls ending with Q and R.
if (last == 0)
dp[p, q, r, last] = countWays(p - 1, q, r, 1) +
countWays(p - 1, q, r, 2);
// Same as above case for 'q' and 'r'
else if (last == 1)
dp[p, q, r, last] = countWays(p, q - 1, r, 0) +
countWays(p, q - 1, r, 2);
//(last==2)
else
dp[p, q, r, last] = countWays(p, q, r - 1, 0) +
countWays(p, q, r - 1, 1);
return dp[p, q, r, last];
}
// Returns count of required arrangements
static int countUtil(int p, int q, int r)
{
// Initialize 'dp' array
for(int i = 0; i < MAX; i++)
for(int j = 0; j < MAX; j++)
for(int k = 0; k < MAX; k++)
for(int l = 0; l < 4; l++)
dp[i, j, k, l] = -1;
// Three cases arise:
return countWays(p, q, r, 0) + // Last required balls is type P
countWays(p, q, r, 1) + // Last required balls is type Q
countWays(p, q, r, 2); // Last required balls is type R
}
// Driver code
static void Main()
{
int p = 1, q = 1, r = 1;
Console.WriteLine(countUtil(p, q, r));
}
}
// This code is contributed by mits.
Python3
# Python3 program to count number of ways to
# arrange three types of balls such that no
# two balls of same color are adjacent to each other
MAX = 100;
# table to store to store results of subproblems
dp = [[[[-1] * 4 for i in range(MAX)]
for j in range(MAX)]
for k in range(MAX)];
# Returns count of arrangements where last
# placed ball is 'last'. 'last' is 0 for 'p',
# 1 for 'q' and 2 for 'r'
def countWays(p, q, r, last):
# if number of balls of any color becomes less
# than 0 the number of ways arrangements is 0.
if (p < 0 or q < 0 or r < 0):
return 0;
# If last ball required is of type P and the
# number of balls of P type is 1 while number
# of balls of other color is 0 the number of
# ways is 1.
if (p == 1 and q == 0 and
r == 0 and last == 0):
return 1;
# Same case as above for 'q' and 'r'
if (p == 0 and q == 1 and
r == 0 and last == 1):
return 1;
if (p == 0 and q == 0 and
r == 1 and last == 2):
return 1;
# If this subproblem is already evaluated
if (dp[p][q][r][last] != -1):
return dp[p][q][r][last];
# if last ball required is P and the number
# of ways is the sum of number of ways to
# form sequence with 'p-1' P balls, q Q Balls
# and r R balls ending with Q and R.
if (last == 0):
dp[p][q][r][last] = (countWays(p - 1, q, r, 1) +
countWays(p - 1, q, r, 2));
# Same as above case for 'q' and 'r'
elif (last == 1):
dp[p][q][r][last] = (countWays(p, q - 1, r, 0) +
countWays(p, q - 1, r, 2));
else:
#(last==2)
dp[p][q][r][last] = (countWays(p, q, r - 1, 0) +
countWays(p, q, r - 1, 1));
return dp[p][q][r][last];
# Returns count of required arrangements
def countUtil(p, q, r):
# Three cases arise:
# Last required balls is type P
# Last required balls is type Q
# Last required balls is type R
return (countWays(p, q, r, 0) +
countWays(p, q, r, 1) +
countWays(p, q, r, 2));
# Driver Code
p, q, r = 1, 1, 1;
print(countUtil(p, q, r));
# This code is contributed by mits
PHP
Javascript
输出:
6
该解决方案的时间复杂度是指数级的。
我们可以观察到,有许多子问题一次又一次地被解决,因此可以使用动态规划(DP)来解决问题。我们可以轻松地对这个问题进行记忆化解决方案。
C++
// C++ program to count number of ways to arrange three
// types of balls such that no two balls of same color
// are adjacent to each other
#include
using namespace std;
#define MAX 100
// table to store to store results of subproblems
int dp[MAX][MAX][MAX][3];
// Returns count of arrangements where last placed ball is
// 'last'. 'last' is 0 for 'p', 1 for 'q' and 2 for 'r'
int countWays(int p, int q, int r, int last)
{
// if number of balls of any color becomes less
// than 0 the number of ways arrangements is 0.
if (p<0 || q<0 || r<0)
return 0;
// If last ball required is of type P and the number
// of balls of P type is 1 while number of balls of
// other color is 0 the number of ways is 1.
if (p==1 && q==0 && r==0 && last==0)
return 1;
// Same case as above for 'q' and 'r'
if (p==0 && q==1 && r==0 && last==1)
return 1;
if (p==0 && q==0 && r==1 && last==2)
return 1;
// If this subproblem is already evaluated
if (dp[p][q][r][last] != -1)
return dp[p][q][r][last];
// if last ball required is P and the number of ways is
// the sum of number of ways to form sequence with 'p-1' P
// balls, q Q Balls and r R balls ending with Q and R.
if (last==0)
dp[p][q][r][last] = countWays(p-1,q,r,1) + countWays(p-1,q,r,2);
// Same as above case for 'q' and 'r'
else if (last==1)
dp[p][q][r][last] = countWays(p,q-1,r,0) + countWays(p,q-1,r,2);
else //(last==2)
dp[p][q][r][last] = countWays(p,q,r-1,0) + countWays(p,q,r-1,1);
return dp[p][q][r][last];
}
// Returns count of required arrangements
int countUtil(int p, int q, int r)
{
// Initialize 'dp' array
memset(dp, -1, sizeof(dp));
// Three cases arise:
return countWays(p, q, r, 0) + // Last required balls is type P
countWays(p, q, r, 1) + // Last required balls is type Q
countWays(p, q, r, 2); // Last required balls is type R
}
// Driver code to test above
int main()
{
int p = 1, q = 1, r = 1;
printf("%d", countUtil(p, q, r));
return 0;
}
Java
// Java program to count number
// of ways to arrange three
// types of balls such that no
// two balls of same color
// are adjacent to each other
import java.util.Arrays;
class GFG
{
static final int MAX = 100;
// table to store to store results of subproblems
static int dp[][][][] = new int[MAX][MAX][MAX][3];
// Returns count of arrangements
// where last placed ball is
// 'last'. 'last' is 0 for 'p',
// 1 for 'q' and 2 for 'r'
static int countWays(int p, int q, int r, int last)
{
// if number of balls of any
// color becomes less than 0
// the number of ways arrangements is 0.
if (p < 0 || q < 0 || r < 0)
return 0;
// If last ball required is
// of type P and the number
// of balls of P type is 1
// while number of balls of
// other color is 0 the number
// of ways is 1.
if (p == 1 && q == 0 && r == 0 && last == 0)
return 1;
// Same case as above for 'q' and 'r'
if (p == 0 && q == 1 && r == 0 && last == 1)
return 1;
if (p == 0 && q == 0 && r == 1 && last == 2)
return 1;
// If this subproblem is already evaluated
if (dp[p][q][r][last] != -1)
return dp[p][q][r][last];
// if last ball required is P and
// the number of ways is the sum
// of number of ways to form sequence
// with 'p-1' P balls, q Q balss and
// r R balls ending with Q and R.
if (last == 0)
dp[p][q][r][last] = countWays(p - 1, q, r, 1) +
countWays(p - 1, q, r, 2);
// Same as above case for 'q' and 'r'
else if (last == 1)
dp[p][q][r][last] = countWays(p, q - 1, r, 0) +
countWays(p, q - 1, r, 2);
//(last==2)
else
dp[p][q][r][last] = countWays(p, q, r - 1, 0) +
countWays(p, q, r - 1, 1);
return dp[p][q][r][last];
}
// Returns count of required arrangements
static int countUtil(int p, int q, int r)
{
// Initialize 'dp' array
for (int[][][] row : dp)
{
for (int[][] innerRow : row)
{
for (int[] innerInnerRow : innerRow)
{
Arrays.fill(innerInnerRow, -1);
}
}
};
// Three cases arise:
return countWays(p, q, r, 0) + // Last required balls is type P
countWays(p, q, r, 1) + // Last required balls is type Q
countWays(p, q, r, 2); // Last required balls is type R
}
// Driver code
public static void main(String[] args)
{
int p = 1, q = 1, r = 1;
System.out.print(countUtil(p, q, r));
}
}
// This code is contributed by Anant Agarwal.
C#
// C# program to count number
// of ways to arrange three
// types of balls such that no
// two balls of same color
// are adjacent to each other
using System;
class GFG
{
static int MAX = 101;
// table to store to store results of subproblems
static int[,,,] dp = new int[MAX, MAX, MAX, 4];
// Returns count of arrangements
// where last placed ball is
// 'last'. 'last' is 0 for 'p',
// 1 for 'q' and 2 for 'r'
static int countWays(int p, int q, int r, int last)
{
// if number of balls of any
// color becomes less than 0
// the number of ways arrangements is 0.
if (p < 0 || q < 0 || r < 0)
return 0;
// If last ball required is
// of type P and the number
// of balls of P type is 1
// while number of balls of
// other color is 0 the number
// of ways is 1.
if (p == 1 && q == 0 && r == 0 && last == 0)
return 1;
// Same case as above for 'q' and 'r'
if (p == 0 && q == 1 && r == 0 && last == 1)
return 1;
if (p == 0 && q == 0 && r == 1 && last == 2)
return 1;
// If this subproblem is already evaluated
if (dp[p, q, r, last] != -1)
return dp[p, q, r, last];
// if last ball required is P and
// the number of ways is the sum
// of number of ways to form sequence
// with 'p-1' P balls, q Q balss and
// r R balls ending with Q and R.
if (last == 0)
dp[p, q, r, last] = countWays(p - 1, q, r, 1) +
countWays(p - 1, q, r, 2);
// Same as above case for 'q' and 'r'
else if (last == 1)
dp[p, q, r, last] = countWays(p, q - 1, r, 0) +
countWays(p, q - 1, r, 2);
//(last==2)
else
dp[p, q, r, last] = countWays(p, q, r - 1, 0) +
countWays(p, q, r - 1, 1);
return dp[p, q, r, last];
}
// Returns count of required arrangements
static int countUtil(int p, int q, int r)
{
// Initialize 'dp' array
for(int i = 0; i < MAX; i++)
for(int j = 0; j < MAX; j++)
for(int k = 0; k < MAX; k++)
for(int l = 0; l < 4; l++)
dp[i, j, k, l] = -1;
// Three cases arise:
return countWays(p, q, r, 0) + // Last required balls is type P
countWays(p, q, r, 1) + // Last required balls is type Q
countWays(p, q, r, 2); // Last required balls is type R
}
// Driver code
static void Main()
{
int p = 1, q = 1, r = 1;
Console.WriteLine(countUtil(p, q, r));
}
}
// This code is contributed by mits.
蟒蛇3
# Python3 program to count number of ways to
# arrange three types of balls such that no
# two balls of same color are adjacent to each other
MAX = 100;
# table to store to store results of subproblems
dp = [[[[-1] * 4 for i in range(MAX)]
for j in range(MAX)]
for k in range(MAX)];
# Returns count of arrangements where last
# placed ball is 'last'. 'last' is 0 for 'p',
# 1 for 'q' and 2 for 'r'
def countWays(p, q, r, last):
# if number of balls of any color becomes less
# than 0 the number of ways arrangements is 0.
if (p < 0 or q < 0 or r < 0):
return 0;
# If last ball required is of type P and the
# number of balls of P type is 1 while number
# of balls of other color is 0 the number of
# ways is 1.
if (p == 1 and q == 0 and
r == 0 and last == 0):
return 1;
# Same case as above for 'q' and 'r'
if (p == 0 and q == 1 and
r == 0 and last == 1):
return 1;
if (p == 0 and q == 0 and
r == 1 and last == 2):
return 1;
# If this subproblem is already evaluated
if (dp[p][q][r][last] != -1):
return dp[p][q][r][last];
# if last ball required is P and the number
# of ways is the sum of number of ways to
# form sequence with 'p-1' P balls, q Q Balls
# and r R balls ending with Q and R.
if (last == 0):
dp[p][q][r][last] = (countWays(p - 1, q, r, 1) +
countWays(p - 1, q, r, 2));
# Same as above case for 'q' and 'r'
elif (last == 1):
dp[p][q][r][last] = (countWays(p, q - 1, r, 0) +
countWays(p, q - 1, r, 2));
else:
#(last==2)
dp[p][q][r][last] = (countWays(p, q, r - 1, 0) +
countWays(p, q, r - 1, 1));
return dp[p][q][r][last];
# Returns count of required arrangements
def countUtil(p, q, r):
# Three cases arise:
# Last required balls is type P
# Last required balls is type Q
# Last required balls is type R
return (countWays(p, q, r, 0) +
countWays(p, q, r, 1) +
countWays(p, q, r, 2));
# Driver Code
p, q, r = 1, 1, 1;
print(countUtil(p, q, r));
# This code is contributed by mits
PHP
Javascript
输出:
6
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