• Exchange operates under a perfect economic simulation where each POW token grants you a stake of all cryptocurrency volume. POWX is a loyalty/transitional token in POW Ecosystem, that enables ...
• ## D. Pow

2021-02-07 21:47:14
As we all now that there’s a easy way to describe some same consecutive factors multiply each other which is POW. For example, 2×2×2=232\times2\times2=2^32×2×2=23 and 3×3=323\times3=3^23×3=32 ...
题干
form QLU p1010 D. Pow Description
As we all now that there’s a easy way to describe some same consecutive factors multiply each other which is POW.
For example, 2×2×2=232\times2\times2=2^32×2×2=23 and 3×3=323\times3=3^23×3=32 and it is easy to see that 23<3223<3223<32.
Now, your math teacher just gave you a problem in his class which is 10099100^{99}10099 9910099^{100}99100 which one is bigger? There are many ways to solve that and you solved it quickly. When you told the answer to your teach, he seemed to be unsatisfactory about that and he continued to asked the pow problems like that, which is given four positive integers a,b,c,da,b,c,da,b,c,d could you tell that which one is bigger between aba^bab and cdc^dcd or they are equal. 16121439447368.png
picture: some answers in the Internet In C, you may use the function pow to calculate the answer:
pow(2,3)=8 pow(3,2)=9
Input
First line of the input contains one positive integer T(1≤T≤105)T (1\le T\le 10^5)T(1≤T≤105) indicating the number of the test cases.
Next TTT lines, each line contains four positive integers a,b,c,d(1≤a,b,c,d≤109)a,b,c,d (1\le a,b,c,d\le10^9)a,b,c,d(1≤a,b,c,d≤109) Output
Four each test case: if ab>cdab>cdab>cd print LEFT else if ab=cdab=cdab=cd print EQUAL else print RIGHT Examples Input 复制
3 3 2 2 3 100 99 99 100 2 4 4 2
Output 复制
LEFT RIGHT EQUAL
代码
#include<iostream>
#include<stdio.h>
#include<math.h>
using namespace std;
int main(){
int t;
double a,b,c,d;
cin>>t;
while (t--)
{
scanf("%lf %lf %lf %lf",&a,&b,&c,&d);
double lf=b*log(a),ri=d*log(c);
if(lf-ri>1e-8) printf("LEFT\n");//精度控制在1e-8之间
else if(fabs(lf-ri)<=1e-8) printf("EQUAL\n");
else printf("RIGHT\n");

/* co
de */
}

system("pause");
return 0;
}

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• python中pow函数 Python pow()函数 (Python pow() function) pow() function is a library function in Python, is used to get the x to the power of y, where x is the base and y is the power – in other ...

python中pow函数
Python pow()函数 (Python pow() function)
pow() function is a library function in Python, is used to get the x to the power of y, where x is the base and y is the power – in other words we can say that pow() function calculates the power i.e. x**y – it means x raised to the power y.
pow()函数是Python中的一个库函数，用于将x赋予y的幂，其中x是基数， y是幂–换句话说，我们可以说pow()函数计算了幂，即x ** y –表示x升为y的幂。
Syntax:
句法：
    pow(x, y [,z])


Parameter(s):
参数：
x – A base number which is to be powered  x –要加电的基数 y – A value of the power y –幂的值 z – It's an optional parameter, it is used to define/fine the modules of the result of (x**y). z –这是一个可选参数，用于定义/优化(x ** y)结果的模块。
Return value: numer – returns result.
返回值： numer –返回结果。
Example:
例：
    Input:
x = 2   # base
y = 3   # power
z = 3   # value for modulus

print(pow(x, y))
print(pow(x, y, z))

Output:
8
2


Note:
注意：
pow() function with two arguments (x,y) – it is equivalent to x**y 具有两个参数(x，y)的 pow()函数 –等效于x ** y pow() function with three arguments (x,y,z) – it is equivalent to (x**y) % z 具有三个参数(x，y，z)的 pow()函数 –等效于(x ** y)％z
pow() function results with different types of the values, consider the below given table,
pow()函数结果具有不同类型的值，请考虑以下给定的表，
XYZNegative or Non Negative IntegerNon-negative IntegerMay or may not be presentNegative or Non Negative IntegerNegative IntegerShould not be present
X  ÿ  ž  负整数或非负整数  非负整数  可能存在或可能不存在  负整数或非负整数  负整数  不应该出现
Example1:
范例1：
# python code to demonstrate example of
# pow() function

x = 2   # base
y = 3   # power
z = 3   # value for modulus

# calcilating power with two arguments
result1 = pow(x, y)
# calcilating power & modulus with three arguments
result2 = pow(x, y, z)

# printing the values
print("result1: ", result1)
print("result2: ", result2)


Output
输出量
result1:  8
result2:  2


Note:  pow() can return integer and float both values depend on the condition/ values.
注意： pow()可以返回整数，并且两个浮点数都取决于条件/值。
Example2:
范例2：
# python code to demonstrate example of
# pow() function

x = 4   # base
y = 3   # power
z = 6   # value for modulus

print("With 2 args:", pow(x,y))     #first taking 2 args only
print("With 3 args:", pow(x,y,z))   #then all the 3 args

print("Return float values:", pow(2,-3))

print('Random numbers power:' , pow(5,5,6))


Output
输出量
With 2 args: 64
With 3 args: 4
Return float values: 0.125
Random numbers power: 5


计算任何数量幂的不同方法 (Different approaches to calculate the power of any number)
By using simple approach: (x**y) 通过使用简单的方法：(x ** y) By using pow() function: pow(x,y) 通过使用pow()函数：pow(x，y) By using math.pow() function – which is a function of "math" library 通过使用math.pow()函数 –这是“数学”库的函数
Note:  pow() function takes three arguments (the last one is optional), where math.pow() takes only two arguments. In this, there is no third parameter present else all the functionality is the same as simple pow(). But the math.pow() always returns float values. So if you, for some reason, want to make sure you get float as a result back, then math.pow() will provide that benefit to the user.
注意： pow()函数带有三个参数(最后一个是可选的)，其中math.pow()仅带有两个参数。 在此，不存在第三个参数，否则所有功能都与simple pow()相同。 但是math.pow()始终返回浮点值。 因此，如果由于某种原因要确保返回结果为float，则math.pow()将为用户提供这一好处。
Example 1: Calculating X to the power Y using different approaches
示例1：使用不同的方法将X乘以Y
# python code to demonstrate different
# approaches to calculate x to the power y

import math

x = 2   # base
y = 3   # power

result1 = x**y
result2 = pow(x,y)
result3 = math.pow(x,y)

print("result1 (normal approach): ", result1)
print("result2 (using pow() function): ", result2)
print("result3 (using math.pow() function): ", result3)


Output
输出量
result1 (normal approach):  8
result2 (using pow() function):  8
result3 (using math.pow() function):  8.0


Example2: pow() vs math.pow() with third parameter
示例2：带有第三个参数的pow()vs math.pow()
# python code to demonstrate different
# approaches to calculate x to the power y

import math

x = 2   # base
y = 3   # power
z = 3   # for moduls

print("pow(x,y,z): ", pow(x,y,z))

# following statement will throw an error
print("math.pow(x,y,z): ", math.pow(x,y,z))


Output
输出量
pow(x,y,z):  2
Traceback (most recent call last):
File "/home/main.py", line 12, in

print("math.pow(x,y,z): ", math.pow(x,y,z))
TypeError: pow expected 2 arguments, got 3



翻译自: https://www.includehelp.com/python/pow-function-with-example-in-python.aspx

python中pow函数

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• I would like to ask time complexity of the following code.... (Is time complexity of Math.pow() O(1)? ) In general, is Math.pow(a,b) has time complexity O(b) or O(1)? Thanks in advance.publ...


I would like to ask time complexity of the following code. Is it O(n)? (Is time complexity of Math.pow() O(1)? ) In general, is Math.pow(a,b) has time complexity O(b) or O(1)? Thanks in advance.
public void foo(int[] ar) {
int n = ar.length;
int sum = 0;
for(int i = 0; i < n; ++i) {
sum += Math.pow(10,ar[i]);
}
}
解决方案
@Blindy talks about possible approaches that Java could take in implementing pow.
First of all, the general case cannot be repeated multiplication. It won't work for the general case where the exponent is not an integer. (The signature for pow is Math.pow(double, double)!)
In the OpenJDK 8 codebase, the native code implementation for pow can work in two ways:
The first implementation in e_pow.c uses a power series. The approach is described in the C comments as follows:
* Method: Let x = 2 * (1+f)
* 1. Compute and return log2(x) in two pieces:
* log2(x) = w1 + w2,
* where w1 has 53-24 = 29 bit trailing zeros.
* 2. Perform y*log2(x) = n+y' by simulating multi-precision
* arithmetic, where |y'|<=0.5.
* 3. Return x**y = 2**n*exp(y'*log2)
The second implementation in w_pow.c is a wrapper for the pow function provided by the Standard C library. The wrapper deals with edge cases.
Now it is possible that the Standard C library uses CPU specific math instructions. If it did, and the JDK build (or runtime) selected1 the second implementation, then Java would use those instructions too.
But either way, I can see no trace of any special case code that uses repeated multiplication. You can safely assume that it is O(1).
1 - I haven't delved into how when the selection is / can be made.

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• Which one is more efficient using math.pow or the ** operator? When should I use one over the ...So far I know that x**y can return an int or a float if you use a decimalthe function pow will retur...


Which one is more efficient using math.pow or the ** operator? When should I use one over the other?
So far I know that x**y can return an int or a float if you use a decimal
the function pow will return a float
import math
print math.pow(10, 2)
print 10. ** 2
解决方案
Using the power operator ** will be faster as it won’t have the overhead of a function call. You can see this if you disassemble the Python code:
>>> dis.dis('7. ** i')
6 BINARY_POWER
7 RETURN_VALUE
>>> dis.dis('pow(7., i)')
9 CALL_FUNCTION 2 (2 positional, 0 keyword pair)
12 RETURN_VALUE
>>> dis.dis('math.pow(7, i)')
12 CALL_FUNCTION 2 (2 positional, 0 keyword pair)
15 RETURN_VALUE
Note that I’m using a variable i as the exponent here because constant expressions like 7. ** 5 are actually evaluated at compile time.
Now, in practice, this difference does not matter that much, as you can see when timing it:
>>> from timeit import timeit
>>> timeit('7. ** i', setup='i = 5')
0.2894785532627111
>>> timeit('pow(7., i)', setup='i = 5')
0.41218495570683444
>>> timeit('math.pow(7, i)', setup='import math; i = 5')
0.5655053168791255
So, while pow and math.pow are about twice as slow, they are still fast enough to not care much. Unless you can actually identify the exponentiation as a bottleneck, there won’t be a reason to choose one method over the other if clarity decreases. This especially applies since pow offers an integrated modulo operation for example.
timeit shows that math.pow is slower than ** in all cases. What is math.pow() good for anyway? Has anybody an idea where it can be of any advantage then?
The big difference of math.pow to both the builtin pow and the power operator ** is that it always uses float semantics. So if you, for some reason, want to make sure you get a float as a result back, then math.pow will ensure this property.
Let’s think of an example: We have two numbers, i and j, and have no idea if they are floats or integers. But we want to have a float result of i^j. So what options do we have?
We can convert at least one of the arguments to a float and then do i ** j.
We can do i ** j and convert the result to a float (float exponentation is automatically used when either i or j are floats, so the result is the same).
We can use math.pow.
So, let’s test this:
>>> timeit('float(i) ** j', setup='i, j = 7, 5')
0.7610865891750791
>>> timeit('i ** float(j)', setup='i, j = 7, 5')
0.7930400942188385
>>> timeit('float(i ** j)', setup='i, j = 7, 5')
0.8946636625872202
>>> timeit('math.pow(i, j)', setup='import math; i, j = 7, 5')
0.5699394063529439
As you can see, math.pow is actually faster! And if you think about it, the overhead from the function call is also gone now, because in all the other alternatives we have to call float().
In addition, it might be worth to note that the behavior of ** and pow can be overridden by implementing the special __pow__ (and __rpow__) method for custom types. So if you don’t want that (for whatever reason), using math.pow won’t do that.

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• ## pow函数以及math.h的一些坑

千次阅读 多人点赞 2018-12-12 09:47:26
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