例7.7：在口袋中有红，黄，蓝，白黑5种颜色的球若干，每次从口袋中取出3个球，问得到3种不同颜色的球的可能取法，打印出每种组合的3种颜色

最新推荐文章于 2022-11-30 16:29:58 发布

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最新推荐文章于 2022-11-30 16:29:58 发布

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分类专栏： c++程序枚举法谭浩强文章标签： switch enum c++

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本文通过枚举方法，展示了五种颜色小球的无序和有序排列组合过程，使用C++实现，详细解释了如何避免重复组合，适用于初学者理解和实践基本的组合数学原理。

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//五种颜色的小球，枚举:(无序排列组合）
#include
#include
using namespace std;
enum color {red,yellow,blue,white,black}; //声明枚举类型

int main()
{
color pri;
int i, j, k, loop;
int n = 0; //n计数
for (i = red; i <= black; i++)
for (j = red; j <= black; j++)
for (k = red; k <= black; k++)
if ( i != j && i != k && j != k )
{
cout << setw(3) << ++n<<" " ;
for (loop = 1; loop <= 3; loop++)
{
switch (loop)
{
case 1:pri = color(i); break;
case 2:pri = color(j); break;
case 3:pri = color(k); break;
default:break;
}

					switch (pri)
					{
					case red:    cout << setw(8) << "red"<<"  ";      break;
					case yellow: cout << setw(8) << "yellow" << "  "; break;
					case blue:   cout << setw(8) << "blue" << "  ";   break;
					case white:  cout << setw(8) << "white" << "  ";  break;
					case black:  cout << setw(8) << "black" << "  ";  break;
					default: break;
					}
				}
				cout << endl;
			}	
		
cout << endl<<"total:"<<n<<endl;
return 0;

}

//五种颜色的小球，枚举:（有序排列）
#include
#include
using namespace std;

enum color { red, yellow, blue, white, black }; //声明枚举类型

int main()
{
color pri;
int i, j, k, loop;
int n = 0; //n计数
for (i = red; i <= blue; i++)
for (j = i+1; j <= white; j++)
for (k = j+1; k <= black; k++)
{
cout << setw(3) << ++n << " ";
for (loop = 1; loop <= 3; loop++)
{
switch (loop)
{
case 1:pri = color(i); break; //pri是color类型，i是int 类型，必须要强制转换类型！！！
case 2:pri = color(j); break;
case 3:pri = color(k); break;
default:break;
}

					switch (pri)
					{
					case red:    cout << setw(8) << "red" << "  ";      break;
					case yellow: cout << setw(8) << "yellow" << "  "; break;
					case blue:   cout << setw(8) << "blue" << "  ";   break;
					case white:  cout << setw(8) << "white" << "  ";  break;
					case black:  cout << setw(8) << "black" << "  ";  break;
					default: break;
					}
				}
				cout << endl;
			}

cout << endl << "total:" << n << endl;
return 0;

}