If it is known that the stacking sequence of a stack is 1,2,3,..., N, its output sequence is P1, P2, P3,..., PN. If P1 is n, then Pi is n A) I b) n-i C) n-i + 1 D) uncertainty

If it is known that the stacking sequence of a stack is 1,2,3,..., N, its output sequence is P1, P2, P3,..., PN. If P1 is n, then Pi is n A) I b) n-i C) n-i + 1 D) uncertainty

C)n-i+1
Stack arrangement follows the principle of first in first out
Because P1 is n, which is the first number out of the stack, it means that the numbers in the stack before n are not out of the stack, so this order is certain. You can also know that the last number out of the stack must be 1, that is, PN. It is correct to substitute this formula
It's d
Of course, I'm not sure. It's irregular
Let a B be two subsets of the complete set I, then a is really included in the -- condition that B is (CIA) ∪ B = I
Sufficient and unnecessary conditions
As shown in the figure, let P1, P2, P3 , PN, is the vertex of a positive n-polygon inscribed in a circle O, and P is any point on a circle o,
Prove: vector PP1 & # 710; 2 + PP2 & # 710; 2 +... + PPN & # 710; 2 is a fixed value
Vector PPI = vector Po + vector OPI, I = 1,2.., n vector PPI ^ 2 = vector Po ^ 2 + vector OPI ^ 2 + 2 vector Po * vector OPI = 2R ^ 2 + 2 vector Po * vector OPI, I = 1,2,..., n so, vector PP1 & # 710; 2 + PP2 & # 710; 2 +... + PPN & # 710; 2 = 2nr ^ 2 + 2 vector Po * (vector OP1 + op2 +... + OPN) = 2nr ^ 2, where R
Given the set a = {x | log2 x ≤ 2}. B = (- infinity, a), if a is really contained in B, then the value range of real number a is (C, + infinity), where C=——
∵log2 X≤2 ∴0＜x≤4
∵ A is really contained in B ∵ a > 4
∴c=4
The range of a is (0,4), so a is larger than 4, so C ≥ 4
If you are not sure whether C = 4 is true, you can bring in the test by yourself:)
C=3
The reasons are as follows.
∵x∈（0,4），∴A∈（0,4）
A is really contained in B and B ∈ (- ∞, a)
And a ∈ (C, + ∞),
∴a>4，
∴Cmax=3
The reverse order number of P1, P2,. PN is K. find the reverse order number of PN,. P2, P1
I thought about it for a while and made a draft
First of all, we assume that the reverse order numbers of the elements in P1, P2,. PN are T1, T2 TN
That is to say, the reverse number of P1 is T1 (in fact, T1 = 0, which is written as T1 for the convenience of illustration), and the reverse number of P2 is T2 The reverse number of PN is TN
In addition, from the definition of inverse ordinal number, we can know that P1, P2,. PN are different numbers
For the next step of reasoning, we first explain a conclusion, that is, two adjacent numbers are exchanged once. If the former number is larger than the latter number, then the reverse order number of the whole sequence is reduced by 1, otherwise it is increased by 1
That is to say, for the sequence a, B → B, a
If a > b, then the inverse ordinal number will change from 1 to 0, and vice versa
In this way, it is easy to know that for an element PN in the sequence P1, P2,. PN, its reverse order number is TN, which means that in P1, P2,. PN-1, the number of numbers larger than PN is TN, and the number of numbers smaller than PN is n-1-tn
Then P1, P2,. PN transform to PN, P1, P2,. PN-1 needs to go through n-1 times of adjacent transform, and if considered as a whole, there are TN times of inverse number minus 1 transform and n-1-tn times of inverse number plus 1 transform, so the inverse number of PN, P1, P2,. PN-1 is k-tn + n-1-tn
In the same way, for PN-1, for the sequence PN, P1, P2,. PN-1, in PN, P1, P2,. Pn-2, there are TN-1 numbers larger than that and n-2 numbers smaller than that, so the transformation of PN, P1, P2,. PN-1 to PN, PN-1, P1, P2,. Pn-2 needs n-2 times of transformation, and if considered as a whole, it needs to go through the transformation of TN-1 times of inverse number minus 1, and the transformation of n-1-tn-1 times of inverse number plus 1, so PN, PN-1, P1, P2, The reverse order number of. Pn-2 becomes k-tn + n-1-tn-tn-1 + n-2-tn-1
……
And so on, until PN,. P2, P1, we can prove that its inverse ordinal number is k-tn + n-1-tn-tn-1 + n-2-tn-1-tn-2 + n-3-tn-2 - -t1+0-t1=k-2(tn+tn-1+…… +t1)+n-1+n-2+…… +1
We know TN + TN-1 + from the known conditions +t1=k
So the reverse order number of PN,. P2, P1 is k-2k + (n-1) n / 2 = (n-1) n / 2-k
After n examples, it is proved that the process is correct~
C (2, n) - K
Given the set a = {x | log2 (x) ≤ 2}, B = (negative infinity, a), if a contains B, then the value range of real number a is (C, positive infinity), where C is equal to
Log2 (x) ≤ 2, X ≤ 4, a = (- ∞, 4)]. A contained in B: (- ∞, 4] contained in (- ∞, a)
4 ＜ a, [the value range of a: (4, + ∞)]
The total power of the series circuit is equal to the sum of the powers: ptotal = P1 + P2 + P3 + +PN [Formula: p1p2 / (P1 + P2)]
How does this work out
In a series circuit, the current of each consumer is equal, and the sum of the voltages of each consumer is equal to the total voltage
Power of single consumer:
Pn=Un·I
Sum of power of all electrical appliances:
P total = u · I = (U1 + U2 + U3 +...) +Un）·I=U1·I+U2·I+U3·I+…… +Un·I=P1+P2+P3+…… +Pn
According to the conservation of energy, the total power in any circuit is equal to the sum of the load power.
A = {x | log2 (x) ≤ 2}, B = (negative infinity, a). If a is contained in B, then the value range of real number a is (C, positive infinity), C=
Note that a is contained in B, not a contains B
Easy to get a = {x | x ≤ 4}
From the meaning of the title, we can get a > 4
∵a∈(c,+00)
∴c=4
Number prime numbers from small to large: P1 = 2, P2 = 3, P3 = 5 Proving: the nth prime number PN
Bertrand Chebyshev theorem shows that if the integer n > 3, then there is at least one prime P, which conforms to n < p < 2n &; 2
Next, do it yourself
Given the set a = {x | log2 ≤ 2} B = (a, + ∞), if a ∩ B = empty set, then the value range of real number a is
Set a = {x | log2 (x) ≤ 2} = (0,4] B = (a, + ∞)
Let a ∩ B = an empty set, that is, (0,4] ∩ (a, + ∞) = an empty set
a≥4