answer:**Analysis** This question mainly tests the method of converting a point's polar coordinates to Cartesian coordinates, which is a basic problem. By calculating rho from the point's Cartesian coordinates, and then using 1=rhocos theta and sqrt{3}=rhosin theta, we can find theta, and thus determine the polar coordinates of point P. **Solution** Given that the Cartesian coordinates of point P are (1, sqrt{3},3), therefore rho= sqrt{1+3}=2, then from 1=rhocos theta and sqrt{3}=rhosin theta, we can find theta= frac{pi}{3}, thus, the polar coordinates of point P are left(2, frac{pi}{3},3right), therefore, the correct choice is boxed{A}.

answer:Given alpha = k cdot 180^circ + 45^circ (k in mathbb{Z}), - When k is even, the terminal side of k cdot 180^circ is on the positive x-axis, thus alpha = k cdot 180^circ + 45^circ (k in mathbb{Z}) is an angle in the first quadrant. - When k is odd, the terminal side of k cdot 180^circ is on the negative x-axis, thus alpha = k cdot 180^circ + 45^circ (k in mathbb{Z}) is an angle in the third quadrant. Therefore, the terminal side of alpha is in the first or third quadrant. Hence, the correct answer is boxed{text{A}}. This problem involves discussing k as even and odd numbers, initially determining the terminal side of k cdot 180^circ, and adding 45^circ to find the answer. It examines the concept of quadrant angles and axial angles, which is a basic conceptual question and is likely to appear in exams.

answer:1. **Problem Setup**: There are 10 guests, each leaving a unique pair of kaloshes (shoes). Each guest needs to leave by wearing any pair of kaloshes that fit them, which means each guest can only wear kaloshes of their size or larger. At some moment, remaining guests find no kaloshes fitting their feet. 2. **Guests and Kaloshes Numbering**: - Number the guests and kaloshes from 1 to 10 in ascending order of size. - Guest (i) has kaloshes size (i). 3. **Maximal Scenario Proof**: - Suppose there are more than 5 guests left. Specifically, let's assume 6 guests remain. - Since any remaining guest ((6, 7, 8, 9, 10)) can only take kaloshes of size ((6, 7, 8, 9, 10)), the smallest remaining guest (6) can fit the largest remaining kaloshes because sizes are unique and sequential. 4. **Contradiction**: - This leads to a contradiction: if 6 guests remain, the guest with the smallest feet of those remaining (number 6) could always find fitting kaloshes (the largest size among (6, 7, 8, 9, 10)). Hence, it's impossible for more than 5 guests to be left without any kaloshes fitting them. 5. **Possible Remaining Guests**: - If exactly 5 guests have already left, the sizes of the remaining kaloshes would be ((1, 2, 3, 4, 5)). - If guests numbered (1, 2, 3, 4, 5) leave first with kaloshes numbered (10, 9, 8, 7, 6), none of the remaining guests (6, 7, 8, 9, 10) can wear kaloshes (1, 2, 3, 4, 5) (since their feet size is larger). # Conclusion: - In the outlined scenario, the maximum number of guests left unable to find any fit kaloshes is 5. [ boxed{5} ]

answer:Solution: For p, let f(x)=e^{x}-ax-1, f(0)=0. Since exists x > 0, e^{x}-ax < 1 holds, therefore f(x)_{min} < 0. f^{′}(x)=e^{x}-a. It is known that when aleqslant 0, the function f(x) is monotonically increasing, which is discarded. When a > 0, f(x) is monotonically decreasing on (0,ln a) and monotonically increasing on (ln a,+infty). Therefore, when x=ln a, the function f(x) reaches its minimum value, i.e., f(ln a)=a-aln a-1 < 0. Let g(a)=a-aln a-1, g(1)=0. g^{′}(a)=1-ln a-1=-ln a, it is known that when a=1, g(a) reaches its maximum value, thus a > 0 and aneq 1. For q: the function f(x)=-(a-1)^{x} is a decreasing function, then a-1 > 1, solving this yields a > 2. Therefore, p is a necessary but not sufficient condition for q. Hence, the choice is: boxed{B}. By studying the monotonicity of p using derivatives, it can be concluded that a > 0 and aneq 1. For q: the function f(x)=-(a-1)^{x} is a decreasing function, then a-1 > 1, solving this yields a > 2. This allows us to draw the conclusion. This question examines the use of derivatives to study the monotonicity of functions, the monotonicity of composite and exponential functions, and simple logical determination methods, testing reasoning and computational abilities, and is considered a medium-level question.