The statement is as follows: int a [10] = {1,2,3,4,5,6,7,8,9}, * P = a; then the expression with value 6 is () A．*p+6 B．*(p+6) C．p+5 D．*p+=5

The statement is as follows: int a [10] = {1,2,3,4,5,6,7,8,9}, * P = a; then the expression with value 6 is () A．*p+6 B．*(p+6) C．p+5 D．*p+=5

D.
The expression "* P = a" indicates the position where p points to a [0];
Then * P + 6 = a [0] + 6 = 7; a wrong;
If P points to the position of a [0], then p + 6 points to the position of a [6],
Then * (P + 6) = a [6] = 7; B wrong;
P is a pointer, the value of P itself is an address, P + 5 obviously can not be 5; C error;
As mentioned above, P points to the position of a [0], then * P = a [0] = 1;
*P + = 5; equivalent to a [0] + = 5; equivalent to a [0] = a [0] + 5 = 1 + 5 = 6;
Obviously, only D is correct;

For a positive integer with no more than 5 digits, 1 is required to find out how many digits it is, 2 prints out each digit, 3 prints out each digit in reverse order, and C programming language is used

#Include "stdio. H" void main() {unsigned long int a = 0; / / using int may overflow / / int flag = 0; int b [5]; / / the following parts are used to determine the number of digits in the data / / while (flag = = 0) {printf ("input the number ', & A); scanf (% LD", & A); if (a > 0 & & A = 10 & & A = 100)

Input a four digit positive integer x from the keyboard, C language programming to calculate the sum of each number of X, how to write and ideas

#Include "stdio. H" / / contains the header file main() {int x, a, B, C, D; / / declares four variables to store four digits and the number of each digit printf ("please input a four digit integer); / / outputs the prompt statement scanf ("% d ", & x); / / obtains the four digits a = x% 10 from the screen; / / obtains the four bits B = (x% 100

It takes half an hour for Party A and one third an hour for Party B to make the same part. Is that right?

Yes, pro

When the phase difference of two sinusoidal alternating current of the same frequency is equal to 180 degrees, their phase relationship is zero
A: In phase
B: Inverse phase
C: Equal

A. Inverse phase

A school organizes 145 teachers and students to go on an outing. It is known that each car with 45 seats costs 720 yuan, and each car with 30 seats costs 580 yuan
Kuai answers 300 fortune correctly

A 45 seat car costs 16 yuan per seat; a 30 seat car costs 19.33 yuan per seat. Obviously, 45 seat cars should be used as much as possible, and it is cheap to have at least empty seats
They all use 45 seat cars, 4 cars, 35 seats, and the cost is 2880 yuan;
All use 30 seat car, need 5, remaining 5 seats, cost 2900 yuan;
If you use three 45 seat cars to pull 135 people, the cost is 2160 yuan; if you use one 30 seat car to pull 10 people, and the remaining 20 seats, the cost is 580 yuan, a total of 2740 yuan;
If you use two cars with 45 seats and pull 90 people, the cost is 1440 yuan; if you use two cars with 30 seats and pull 55 people and the rest five seats, the cost is 1160 yuan, a total of 2600 yuan; at least

Let the square matrix a satisfy the square of a-a-2e = O, prove that a and a + 2E are invertible, and find the inverse of a and a + 2E

The square of a-a-2e = O
So a (A-E) = 2E, a (A-E) / 2 = e, a is reversible, and a is inverse = (A-E) / 2
So the square of a, the square of a [(A-E) / 2] square = E
And the square of a = a + 2E,
So (a + 2e) [(A-E) / 2] square = e
So a + 2E is invertible and inverse = [(A-E) / 2] square

Given that the complex z = x + Yi (x.y ∈ R) satisfies Z-1 = 1, we can find the value range of modulus of complex Z

Z-1 = 1 means that z is on a circle with radius 1 and center point (1,0)
The value range of modulus of complex Z
That is to find the distance range from the point on the circle to the origin
Because the distance from the center of the circle to the origin is 1, r = 1
So the distance - r = 1-1

Class 2, Grade 7, is going on a tour. Students are going to take photos. If a color negative costs 0.6 yuan, one needs to be printed
35 yuan, one for each reservation, no more than 0. 45 yuan, at least how many students participate in the group photo?

At least X students should take the group photo
0.6 + 0.35 ×＜ (or equal to) 0.45 ×
-0.10 ×＜ (or equal to) - 0.6
× > (or equal to) 6
A: there are at least six students taking group photos

Lim x tends to infinity {1 + [2 / (1 + x)]} ^ X

lim（x→∞）(1+1/x)^x=e
Let 2 / (1 + x) = 1 / T, then x = 2t-1
So LIM (x →∞) {1 + [2 / (1 + x)]} ^ x = LIM (t →∞) (1 + 1 / T) ^ (2t-1) = e ^ 2