CodePudding user response:
The building Lord, the number of each ball is unknown?CodePudding user response:
# include "stdafx. H"
#include
#include
using namespace std;
Int main ()
{
Enum color {red, yellow, blue, white, black};//declare enumerated type color
Color pri.//color types of variables pri
Int I, j, k, n=0, loop;//n is the total different color combinations of
For (I=red; i<=black; I++)//when I for a certain color
{
For (j=red; J<=black; J++)//when the j is a color
{
If (I.=j)//if the color of the ball in the first two different
{
For (k=red; k<=black; K++)//only the first two balls of different colors, just need to check the color of the ball in the third
{
If (k!=i& & k!=j)//the color of the ball in the three different
{
N=n + 1;//that the cumulative value of n + 1
Cout
For (loop=1; Loop<=3; Loop++)//successively to three ball processing
{
The switch (loop)//the value of the loop for 1, 2, 3
{
Case 1: pri=color (I); break;//color is casts (I), the pri has a value of I
Case 2: pri=color (j); break;//j
Case 3: pri=color (k); break;//k
Default: break;
}
Switch (pri)//judge pri values, output the corresponding color
{
Case red: cout
Case yellow: cout
Case blue: cout
Case white: cout
Case black: cout
Default: break;
}
}
Cout}
}
}
}
}
Cout<" Total: "& lt;
return 0;
}
CodePudding user response:
The title is wrong! Such as red balls if only less than 3 and greater than or equal to three, the answer must be different, so the problem must be clearCodePudding user response: