不同颜色的球排列

// 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 <iostream> 
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;
    cout << countUtil(p, q, r) << endl;
    return 0;
}