//中级难度算式构造，包含加、减、乘的任意两个，当结果出现负号时，用括号将结果纠正
private static string CMediumFormula(int[] opds, char[] ops)
{
string formula = null;
char op1 = ops[ran.Next(0, 3)];
char op2 = ops[ran.Next(0, 3)];
if (op1 == '-')
{
CompareAndChange(ref opds[0], ref opds[1]);
}
formula = opds[0].ToString() + op1 + opds[1].ToString() + op2 + opds[2].ToString();
if (CalculatorCore.Calculate(formula) < 0)
{
if (op1 == '-')
{
CompareAndChange(ref opds[0], ref opds[1]);
formula = "(" + opds[0].ToString() + op1 + opds[1].ToString() + ")" + op2 + opds[2].ToString();
if (CalculatorCore.Calculate(formula) < 0)
{
formula = opds[2].ToString() + op2 + "(" + opds[0].ToString() + op1 + opds[1].ToString() + ")";
}
}
else
{
CompareAndChange(ref opds[1], ref opds[2]);
formula = opds[0].ToString() + op1 + "(" + opds[1].ToString() + op2 + opds[2].ToString() + ")";
}
}
return(formula);
}
//高级难度算式构造，整数与分数、分数与分数的混合运算
private static string CDifficultFormula(int[] opds, char[] ops, ref string num)
{
string formula = null;
Fraction fraction1;
Fraction fraction2;
char op = ops[ran.Next(0, 3)];
CompareAndChange(ref opds[0], ref opds[1]);
fraction1 = new Fraction(opds[1], opds[0]);
Simplify(ref fraction1);
CompareAndChange(ref opds[2], ref opds[3]);
fraction2 = new Fraction(opds[3], opds[2]);
Simplify(ref fraction2);
if (0 == ran.Next(1, 100) % 2)//余数为0，分数与整数运算，为1，分数与分数运算
{
formula = fraction1.ToString() + op + opds[2].ToString();
num = FractionCalculator(fraction1, op, new Fraction(opds[2], 1)).ToString();
if (CalculatorCore.Calculate(formula) < 0)
{
formula = opds[2].ToString() + op + fraction1.ToString();
num = FractionCalculator(new Fraction(opds[2], 1), op, fraction1).ToString();
}
}
else
{
formula = fraction1.ToString() + op + fraction2;
num = FractionCalculator(fraction1, op, fraction2).ToString();
if (CalculatorCore.Calculate(formula) < 0)
{
formula = fraction2.ToString() + op + fraction1.ToString();
num = FractionCalculator(fraction2, op, fraction1).ToString();
}
}
return(formula);
}
static Random ran = new Random(GetRandomSeed()); //产生不重复的随机数
//生成算式，complexity表示题的难易程度，范围（0、1、2）
public static string CreateFormula(int complexity)
{
string formula;
char[] operators = { '+', '-', '*', '/' };
int[] operands = new int[4];
string num = null;
for (int i = 0; i < operands.Length; i++)
{
operands[i] = ran.Next(1, 100);//构造四个操作数
if (i > 0)
{
while (operands[i] == operands[i - 1])
{
operands[i] = ran.Next(1, 100);//确保操作数不相同
}
}
}
if (0 == complexity)
{
formula = CEasyFormula(operands, operators);
num = CalculatorCore.Calculate(formula).ToString();//计算算式的结果
}
else if (1 == complexity)
{
formula = CMediumFormula(operands, operators);
num = CalculatorCore.Calculate(formula).ToString();//计算算式的结果
}
else
{
formula = CDifficultFormula(operands, operators, ref num);
}
return(formula + "=" + num);
}