21点游戏，求解答（不用if/else==）
本帖最后由 zhan1251 于 2013-02-05 16:28:15 编辑 Question:
The game of Blackjack requires the player to achieve a total of 21 or to get as close to 21 as possible without
exceeding that limit. The cards used to play the game are valued 2 to 10 with one special card worth either 1 or 11 and
that choice is left to the player. Given the value of three cards (1 10) determine the maximum value of the hand (the
three cards in the possession of the player) where the card valued at 1 can be used as either 1 or 11. Display a value of 0 if
there is no way to make the total of the hand less than or equal to 21.
大意翻译：
1. 输入1-10的3个随意整数；
2. “1”可以当“1”用，也可以当“11”用；
3. 结果为3数之和；
4. 如果结果大于21，输出0； 如果结果小于等于21，输出最好结果。

注意： 1. 不能使用if/else, switch...条件表达；
2. 不能使用&&, ||, !, <, >, ==...

Example #1：
Three values: 10 10 10
Best total: 0

Example #2：
Three values: 5 1 5
Best total: 21

Example #3:
Three values: 1 10 1
Best total: 12

Example #4：
Three values: 10 8 1
Best total: 19

Example #5：
Three values: 1 1 1
Best total: 13 游戏 C语言
[解决办法]
15楼啊，我这块砖头可是秦砖啊。
既然这样，我就把完全正确的代码贴出来吧，省得你纠结：

#include <iostream>
using namespace std;

#define LARGEINT999

int ExistAce(int a, int b, int c)
{
int x = (a / 2) * (b / 2) * (c / 2);
return LARGEINT / (LARGEINT + x);
}

int LargeInt(int a, int b)
{
int x1 = (LARGEINT - b) / (LARGEINT - a);
int x2 = 1 - x1;
return a * x1 + b * x2;
}

int RealValue(int a)
{
return a * ((LARGEINT - a) / (LARGEINT - 21));
}

int BestValue(int a, int b, int c)
{
int R1 = a + b + c;
int R2 = R1 + ExistAce(a, b, c) * 10;
int Rx1 = RealValue(R1);
int Rx2 = RealValue(R2);
return LargeInt(Rx1, Rx2);
}

int main()
{
// 以下是基本原理
printf_s("基本原理一：求两个数中比较大的值\n");
printf_s("A=%d, B=%d, x=%d\n", 15, 18, LargeInt(15, 18));
printf_s("A=%d, B=%d, x=%d\n", 20, 16, LargeInt(20, 16));
printf_s("A=%d, B=%d, x=%d\n", 21, 21, LargeInt(21, 21));
printf_s("\n");
printf_s("基本原理二：大于21就废掉（大于21就取值为0）\n"); 
 
printf_s("M=%d, Mx=%d\n", 18, RealValue(18));
printf_s("M=%d, Mx=%d\n", 21, RealValue(21));
printf_s("M=%d, Mx=%d\n", 25, RealValue(25));
printf_s("\n");

// 取五组数来试算
printf_s("实际计算：\n");
int A = 1, B = 5, C = 5;
printf_s("A=%d, B=%d, C=%d, V=%d\n", A, B, C, BestValue(A, B, C));
A = 1, B = 1, C = 5;
printf_s("A=%d, B=%d, C=%d, V=%d\n", A, B, C, BestValue(A, B, C));
A = 1, B = 9, C = 3;
printf_s("A=%d, B=%d, C=%d, V=%d\n", A, B, C, BestValue(A, B, C));
A = 7, B = 6, C = 5;
printf_s("A=%d, B=%d, C=%d, V=%d\n", A, B, C, BestValue(A, B, C));
A = 8, B = 2, C = 1;
printf_s("A=%d, B=%d, C=%d, V=%d\n", A, B, C, BestValue(A, B, C));
A = 5, B = 7, C = 1;
printf_s("A=%d, B=%d, C=%d, V=%d\n", A, B, C, BestValue(A, B, C));
A = 5, B = 1, C = 9;
printf_s("A=%d, B=%d, C=%d, V=%d\n", A, B, C, BestValue(A, B, C));


system("pause");
return 0;
}

基本原理一：求两个数中比较大的值
A=15, B=18, x=18
A=20, B=16, x=20
A=21, B=21, x=21

基本原理二：大于21就废掉（大于21就取值为0）
M=18, Mx=18
M=21, Mx=21
M=25, Mx=0

实际计算：
A=1, B=5, C=5, V=21
A=1, B=1, C=5, V=17
A=1, B=9, C=3, V=13
A=7, B=6, C=5, V=18
A=8, B=2, C=1, V=21
A=5, B=7, C=1, V=13
A=5, B=1, C=9, V=15
请按任意键继续. . .