Python算24点代码，python24点代码,# 算24程序impor

Python算24点代码，python24点代码,# 算24程序impor

# 算24程序import mathPRECISION = 1E-6COUNT_OF_NUMBER = 4NUMBER_TO_BE_CAL = 24g_number = [4, 4, 7, 7]g_expression = ['', '', '', '']for i in range(0, 4):    g_expression[i] = "%d" % g_number[i]print(g_expression)def solve(n):    if(1 == n):        if(math.fabs(NUMBER_TO_BE_CAL - g_number[0]) < PRECISION):            print(g_expression[0])            return True        else:            return False    else:        for i in range(0, n):            for j in range(i+1, n):                a = g_number[i]                b = g_number[j]                #**********************************                #   将剩下的有效数字往前挪，                #   由于两数计算结果保存在number[i]中，                #   所以将数组末元素覆盖number[j]即可                #   *******************************                g_number[j] = g_number[n - 1]                expa = g_expression[i]                expb = g_expression[j]                g_expression[j] = g_expression[n - 1]                # 计算a+b                g_expression[i] = '(' + expa + '+' + expb + ')'                g_number[i] = a + b                if ( solve(n - 1) ) :                    return True;                # 计算a-b                g_expression[i] = '(' + expa + '-' + expb + ')'                g_number[i] = a - b                if ( solve(n - 1) ) :                    return True                # 计算b-a                g_expression[i] = '(' + expb + '-' + expa + ')'                g_number[i] = b - a                if ( solve(n - 1) ):                    return True                # 计算(a*b)                g_expression[i] = '(' + expa + '*' + expb + ')'                g_number[i] = a * b                if ( solve(n - 1) ):                    return True;                # 计算(a/b)                if (b != 0) :                    g_expression[i] = '(' + expa + '/' + expb + ')'                    g_number[i] = a / b                    if ( solve(n - 1) ) :                        return True                # 计算(b/a)                    if (a != 0) :                        g_expression[i] = '(' + expb + '/' + expa + ')'                        g_number[i] = b / a                        if ( solve(n - 1) ):                            return True                 # 恢复现场                g_number[i] = a                g_number[j] = b                g_expression[i] = expa                g_expression[j] = expb        return Falseif(not solve(COUNT_OF_NUMBER)):    print('no solution')#该片段来自于http://byrx.net