Is there a C++ library that can take nth roots of big numbers (numbers than can't fit in an unsigned long long)?
You can use GMP, a popular open source arbitrary-precision math library. It has C++ bindings.
If you want to code this yourself, check out the Wikipedia page on nth roots:
http://en.wikipedia.org/wiki/Nth_root
The iterative algorithm is quite simple:
The nth root of a number A can be computed by the nth root algorithm, a special case of Newton's method. Start with an initial guess x(0) and then iterate using the recurrence relation
x(k+1) = [(n - 1) * x(k) + A / x(k)^(n - 1)] / n
Stop once you've converged to the desired accuracy.
It depends how much bigger than 2^64 you want to go, I guess. Just using doubles is good to about 1 part in 10^9. I wrote a test program in C:
#include <stdio.h>
#include <math.h>
int main(int argc, char **argv)
{
unsigned long long x;
double dx;
int i;
//make x the max possible value
x = ~0ULL;
dx = (double)x;
printf("Starting with dx = %f\n", dx);
//print the 2th to 20th roots
for (i = 2; i < 21; i++)
{
printf("%dth root %.15f\n", i, pow(dx, 1.0/i));
}
return 0;
}
which produced the following output:
Starting with dx = 18446744073709551616.000000
2th root 4294967296.000000000000000
3th root 2642245.949629130773246
4th root 65536.000000000000000
5th root 7131.550214521852467
6th root 1625.498677215435691
7th root 565.293831000991759
8th root 256.000000000000000
9th root 138.247646578215154
10th root 84.448506289465257
11th root 56.421840319745364
12th root 40.317473596635935
13th root 30.338480458853493
14th root 23.775908626191171
15th root 19.248400577313866
16th root 16.000000000000000
17th root 13.592188707483222
18th root 11.757875938204789
19th root 10.327513583579238
20th root 9.189586839976281
Then I compared with Wolfram Alpha for each root to get the error I quoted above.
Depending on your application, perhaps this will be good enough.
Try also MAPM and qd.
MAPM is written in C but also has a C++ API. qd is written in C++ but also has a C API.
Heres a perfect loop for it. I get the exact answer everytime.
// n1 = <input>, n2 = <base limit>, nmrk = <number of cycles>
long double n3 = 0, n2 = 0, n1 = input_number, n4 = 0;
long double mk = 0, tptm = 0, mnk = 0, dad = 0;
for (n3 = 0; tptm != n1 || mnk > 65535 ; nmrk++) {
n4 += 0.19625;
n2 += 0.15625;
n3 += 0.015625;
mk += 0.0073125;
dad += 0.00390625;
mnk = pow(n1, 1.0/(n4+n2+mk+n3+dad));
tptm = pow((mnk), (n4+n2+mk+n3+dad));
}
if (tptm - n1 < 1)
{
uint64_t y = (tptm);
return make_tuple(nmrk, (n1 - y), mnk);
}
I found this to minutes faster
Long division method is the best method to compute the nth root of any positive real number. It gives the best precision of each digit computed. No initial guess and no iterative approximation is required.
Related
Writing a program to solve problem four of project euler: Find the largest palindrome made from the product of two 2-digit numbers. Heres my reprex:
#include <iostream>
int reverseNumber(int testNum)
{
int reversedNum, remainder = 0;
int temp = testNum;
while(temp != 0)
{
remainder = temp % 10;
reversedNum = reversedNum * 10 + remainder;
temp /= 10;
}
return reversedNum;
}
int main()
{
const int MIN = 100;
int numOne = 99;
int product = 0;
for(int numTwo = 10; numTwo < 100; numTwo++)
{
product = numOne * numTwo;
if (reverseNumber(product) == product)
{
int solution = product;
std::cout << solution << '\n';
return 0;
}
}
return 0;
}
My main thought process behind this is that the for loop will go through every number from 10 to 99 and multiply it by 99. My intended outcome is for it to print 9009 which is the largest palindrome with 2 factors of 2 digits. So what I think should happen here is the for loop will go from 10 to 99, and each loop it should go through the parameters of the if statement which reverses the number and sees if it equals itself.
I've made sure it wasn't a compiler issue, as this is recurring between different compilers. The reverseNumber() function returns the proper number every time I've tested it, so that shouldn't be the problem, however this problem only occurs when the function is involved in the logical comparison. By this I mean if that even I set it equal to a variable and put the variable in the if parameters, the issue still occurs. I'm pretty much stumped. I just hope it's not some silly mistake as I've been on this for a couple days now.
int reversedNum, remainder = 0;
You should be aware that this gives you (in an automatic variable context) a zero remainder but an arbitrary reversedNum. This is actually one of the reasons some development shops have the "one variable per declaration" rule.
In other words, it should probably be:
int reversedNum = 0, remainder;
or even:
int reversedNum = 0;
int remainder;
One other thing that often helps out is to limit the scope of variable to as small an area as possible, only bringing them into existence when needed. An example of that would be:
int reverseNumber(int testNum) {
int reversedNum = 0;
while (testNum != 0) {
int remainder = testNum % 10;
reversedNum = reversedNum * 10 + remainder;
testNum /= 10;
}
return reversedNum;
}
In fact, I'd probably go further and eliminate remainder altogether since you only use it once:
reversedNum = reversedNum * 10 + testNum % 10;
You'll notice I've gotten rid of temp there as well. There's little to gain by putting testNum into a temporary variable since it's already a copy of the original (as it was passed in by value).
And one other note, more to do with the problem rather than the code. You seem to be assuming that there is a palindrome formed that is a multiple of 99. That may be the case but a cautious programmer wouldn't rely on it - if you're allowed to assume things like that, you could just replace your entire program with:
print 9009
Hence you should probably check all possibilities.
You also get the first one you find which is not necessarily the highest one (for example, let's assume that 99 * 17 and 99 * 29 are both palindromic - you don't want the first one.
And, since you're checking all possibilities, you probably don't want to stop at the first one, even if the nested loops are decrementing instead of incrementing. That's because, if 99 * 3 and 97 * 97 are both palindromic, you want the highest, not the first.
So a better approach may be to start high and do an exhaustive search, while also ensuring you ignore the palindrome check of candidates that are smaller that your current maximum, something like (pseudo-code)
# Current highest palindrome.
high = -1
# Check in reverse order, to quickly get a relatively high one.
for num1 in 99 .. 0 inclusive:
# Only need to check num2 values <= num1: if there was a
# better palindrome at (num2 * num1), we would have
# already found in with (num1 * num2).
for num2 in num1 .. 0 inclusive:
mult = num1 * num2
# Don't waste time doing palindrome check if it's
# not greater than current maximum - we can't use
# it then anyway. Also, if we find one, it's the
# highest possible for THIS num1 value (since num2
# is decreasing), so we can exit the num2 loop
# right away.
if mult > high:
if mult == reversed(mult):
high = mult
break
if high >= 0:
print "Solution is ", high
else:
print "No solution"
In addition to properly initializing your variables, if you want the largest palindrome, you should switch the direction of your for loop -- like:
for(int numTwo = 100; numTwo > 10; numTwo--) {
...
}
or else you are just printing the first palindrome within your specified range
The statement check is where I don't understand why it shows wrong answer on submission when I write "sum = (solution[R]-solution[L-1])%mod;" instead. Here I have not added mod within the bracket. I don't see how the answer changes by adding a value of taking the mod of same. Problem code in codechef: https://www.codechef.com/problems/FFC219B
#include<iostream>
#define ll long long
#define mod 1000000007 //the modulus we need to take for the final answer
#define endl "\n"
using namespace std;
long long solution[100007] = {0}; //Initialising all the values with zero
int main(){
ios_base :: sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
solution[0] = 0;
ll a1=1,a2=2,a3=3,a4=4; //The variable initialising as per the problem
for(int i = 1;i <= 100007;i++){
ll k=(a1 * a2) % mod * a3 % mod * a4 % mod;
solution[i] = (solution[i-1]+k)%mod; //Adding the previous values as we are to find the sum in range
a1++;
a2++;
a3++;
a4++;
}
int t; //Taking input for number of test cases
cin>>t;
while(t-->0)
{
int L,R;
cin>>L>>R; //Taking the range input
long long sum = 0;
sum = (solution[R]-solution[L-1] + mod)%mod; //statement check & final answer
cout<<sum<<endl;
}
return 0;
}
The program can give the incorrect answer since the correct answer must always be a positive - not a negative - number.
When you subtract consecutive modulo values, the result may well be negative even though the numbers themselves are increasing (eg, (4^3)%10 - (4^2)%10 = 64%10 - 16%10 = 4-6 = -2), . This means “solution[R]-solution[L-1]” may also well be negative, which means “(solution[R]-solution[L-1]) % mod” will also be negative - although clearly the answer (the number of people affected) must always be positive.
So adding the mod value in this fashion ensures that the result will always be positive.
The number is huge (cannot fit in the bounds of unsigned long long int in C++). How do we check?
There is a solution given here but it doesn't make much sense.
The solution here tries to repeatedly divide the large number (represented as a string) by 2 but I'm not sure I understand how the result is reached step by step.
Can someone please explain this or propose a better solution?
We cannot use any external libraries.
This is the sample code:
int isPowerOf2(char* str)
{
int len_str = strlen(str);
// sum stores the intermediate dividend while
// dividing.
int num = 0;
// if the input is "1" then return 0
// because 2^k = 1 where k >= 1 and here k = 0
if (len_str == 1 && str[len_str - 1] == '1')
return 0;
// Divide the number until it gets reduced to 1
// if we are successfully able to reduce the number
// to 1 it means input string is power of two if in
// between an odd number appears at the end it means
// string is not divisible by two hence not a power
// of 2.
while (len_str != 1 || str[len_str - 1] != '1') {
// if the last digit is odd then string is not
// divisible by 2 hence not a power of two
// return 0.
if ((str[len_str - 1] - '0') % 2 == 1)
return 0;
// divide the whole string by 2. i is used to
// track index in current number. j is used to
// track index for next iteration.
for (int i = 0, j = 0; i < len_str; i++) {
num = num * 10 + str[i] - '0';
// if num < 2 then we have to take another digit
// to the right of A[i] to make it bigger than
// A[i]. E. g. 214 / 2 --> 107
if (num < 2) {
// if it's not the first index. E.g 214
// then we have to include 0.
if (i != 0)
str[j++] = '0';
// for eg. "124" we will not write 064
// so if it is the first index just ignore
continue;
}
str[j++] = (int)(num / 2) + '0';
num = (num) - (num / 2) * 2;
}
str[j] = '\0';
// After every division by 2 the
// length of string is changed.
len_str = j;
}
// if the string reaches to 1 then the str is
// a power of 2.
return 1;
}
I'm trying to understand the process in the while loop. I know there are comments but they arent really helping me glean through the logic.
Let's start by figuring out how to halve a "string-number". We'll start with 128 as an example. You can halve each digit in turn (starting from the left), keeping in mind that an odd number affects the digit on the right(a). So, for the 1 in 128, you halve that to get zero but, because it was odd, five should be kept in storage to be added to the digit on its right (once halved):
128
v
028
Then halve the 2 as follows (adding back in that stored 5):
028
v
018
v
068
Because that wasn't odd, we don't store a 5 for the next digit so we halve the 8 as follows:
068
v
064
You can also make things easier then by stripping off any leading zeros. From that, you can see that it correctly halves 128 to get 64.
To see if a number is a power of two, you simply keep halving it until you reach exactly 1. But, if at any point you end up with an odd number (something ending with a digit from {1, 3, 5, 7, 9}, provided it's not the single-digit 1), it is not a power of two.
By way of example, the following Python 3 code illustrates the concept:
import re, sys
# Halve a numeric string. The addition of five is done by
# Choosing the digit from a specific set (lower or upper
# digits).
def half(s):
halfS = '' # Construct half value.
charSet = '01234' # Initially lower.
for digit in s: # Digits left to right.
if digit in '13579': # Select upper for next if odd.
nextCharSet = '56789'
else:
nextCharSet = '01234' # Otherwise lower set.
halfS += charSet[int(digit) // 2] # Append half value.
charSet = nextCharSet # And prep for next digit.
while halfS[0] == '0': # Remove leading zeros.
halfS = halfS[1:]
return halfS
# Checks for validity.
if len(sys.argv) != 2:
print('Needs a single argument')
sys.exit(1)
num = sys.argv[1]
if not re.match('[1-9][0-9]*', num):
print('Argument must be all digits')
sys.exit(1)
print(num)
while num != '1':
if num[-1:] in '13579':
print('Reached odd number, therefore cannot be power of two')
sys.exit(0)
num = half(num)
print(num)
print('Reached 1, was therefore power of two')
Running that with various (numeric) arguments will show you the process, such as with:
pax$ python ispower2.py 65534
65534
32767
Reached odd number, therefore cannot be power of two
pax$ python ispower2.py 65536
65536
32768
16384
8192
4096
2048
1024
512
256
128
64
32
16
8
4
2
1
Reached 1, was therefore power of two
(a) Take, for example, the number 34. Half of the 3 is 1.5 so the 1 can be used to affect that specific digit position but the "half" left over can simply be used by bumping up the digit on the right by five after halving it. So the 4 halves to a 2 then has five added to make 7. And half of 34 is indeed 17.
This solution does work only for numbers which are not too large i.e. fits in the range of unsigned long long int.
Simpler C++ solution using bitmanipulation for small numbers :-
int power(string s) {
// convert number to unsigned long long int
// datatype can be changed to long int, int as per the requirement
// we can also use inbuilt function like stol() or stoll() for this
unsigned long long int len = s.length();
unsigned long long int num = s[0]-'0';
for(unsigned long long int i = 1; i<len; i++)
num = (num*10)+(s[i]-'0');
if(num == 1)
return 0;
//The powers of 2 have only one set bit in their Binary representation
//If we subtract 1 from a power of 2 what we get is 1s till the last unset bit and if we apply Bitwise AND operator we should get only zeros
if((num & (num-1)) == 0)
return 1;
return 0;
}
A bit better solution that I could code in Java, which doesn't use any fancy object like BigInteger. This approach is same as simple way of performing division. Only look out for remainder after each division. Also trim out the leading zeroes from the quotient which becomes new dividend for next iteration.
class DivisionResult{
String quotient;
int remainder;
public DivisionResult(String q, int rem){
this.quotient = q;
this.remainder = rem;
}
}
public int power(String A) {
if (A.equals("0") || A.equals("1")) return 0;
while (!A.equals("1")){
DivisionResult dr = divideByTwo(A);
if (dr.remainder == 1) return 0;
A = dr.quotient;
}
return 1;
}
public DivisionResult divideByTwo(String num){
StringBuilder sb = new StringBuilder();
int carry = 0;
for (int i = 0;i < num.length(); i++){
int divisibleNum = carry*10 + (num.charAt(i) - '0');
carry = divisibleNum%2;
sb.append(divisibleNum/2);
}
return new DivisionResult(sb.toString().replaceAll("^0+(?!$)", ""), carry);
}
I know how to obtain the square root of a number using the sqrt function.
How can I obtain the cube root of a number?
sqrt stands for "square root", and "square root" means raising to the power of 1/2. There is no such thing as "square root with root 2", or "square root with root 3". For other roots, you change the first word; in your case, you are seeking how to perform cube rooting.
Before C++11, there is no specific function for this, but you can go back to first principles:
Square root: std::pow(n, 1/2.) (or std::sqrt(n))
Cube root: std::pow(n, 1/3.) (or std::cbrt(n) since C++11)
Fourth root: std::pow(n, 1/4.)
etc.
If you're expecting to pass negative values for n, avoid the std::pow solution — it doesn't support negative inputs with fractional exponents, and this is why std::cbrt was added:
std::cout << std::pow(-8, 1/3.) << '\n'; // Output: -nan
std::cout << std::cbrt(-8) << '\n'; // Output: -2
N.B. That . is really important, because otherwise 1/3 uses integer division and results in 0.
in C++11 std::cbrt was introduced as part of math library, you may refer
include <cmath>
std::pow(n, 1./3.)
Also, in C++11 there is cbrt in the same header.
Math for Dummies.
The nth root of x is equal to x^(1/n), so use std::pow. But I don't see what this has to with operator overloading.
Just to point this out, though we can use both ways but
long long res = pow(1e9, 1.0/3);
long long res2 = cbrt(1e9);
cout<<res<<endl;
cout<<res2<<endl;
returns
999
1000
So, in order to get the correct results with pow function we need to add an offset of 0.5 with the actual number or use a double data type i.e.
long long res = pow(1e9+0.5, 1.0/3)
double res = pow(1e9, 1.0/3)
more detailed explanation here C++ pow unusual type conversion
Actually the round must go for the above solutions to work.
The Correct solution would be
ans = round(pow(n, 1./3.));
The solution for this problem is
cube_root = pow(n,(float)1/3);
and you should #include <math.h> library file
Older standards of C/C++ don't support cbrt() function.
When we write code like cube_root = pow(n,1/3); the compiler thinks 1/3 = 0 (division problem in C/C++), so you need to do typecasting using (float)1/3 in order to get the correct answer
#include<iostream.h>
#include<conio.h>
#include<math.h>
using namespace std;
int main(){
float n = 64 , cube_root ;
clrscr();
cube_root = pow(n , (float)1/3);
cout<<"cube root = "<<cube_root<<endl;
getch();
return 0;
}
cube root = 4
You can try this C algorithm :
// return a number that, when multiplied by itself twice, makes N.
unsigned cube_root(unsigned n){
unsigned a = 0, b;
for (int c = sizeof(unsigned) * CHAR_BIT / 3 * 3 ; c >= 0; c -= 3) {
a <<= 1;
b = 3 * a * (a + 1) + 1;
if (n >> c >= b)
n -= b << c, ++a;
}
return a;
}
I would discourage any of the above methods as they didn't work for me. I did pow(64, 1/3.) along with pow(64, 1./3.) but the answer I got was 3
Here's my logic.
ans = pow(n, 1/3.);
if (pow(ans, 3) != n){
ans++;
}
this works for me but I don't understand how it works at all. Could anyone explain?
for(int round = 0; round < rounds_count; round++)
{
for(int match = 0; match < matches_per_round; match++)
{
int home = (round + match) % (teams_count - 1);
int away = (teams_count - 1 - match + round) % (teams_count - 1);
if(match == 0)
away = teams_count - 1;
matches.push_back(Match(&teams[home], &teams[away], round));
}
}
What's the trick with modulo?
I'm not sure why this would be using teams_count-1 instead of teams_count, but in general, the modulus is making it "wrap around" so that if round+match is greater than the last team number, it will wrap back ground to one of the first teams instead of going past the last team.
The way away is handled, is a bit special. The % operator doesn't wrap around the way you want when you have negative numbers. For example -1 % 5 gives you -1 instead of 4. A trick to get around this problem is to add your divisor. (-1+5)%5 gives you 4.
Let's rework the code a little to make it clearer. First I'll use another variable n to represent the number of teams (again I'm not sure why teams_count-1 is used for this in your code):
int n = teams_count-1;
int home = (round + match) % n;
int away = (n - match + round) % n;
Then I'll reorganize the away calculation a little:
int n = teams_count-1;
int home = (round + match) % n;
int away = (round - match + n) % n;
It should now be clearer that the home team is starting with the current round and then adding the match, while the away team is starting with the current round and subtracting the match. The % n makes it wrap around, and the + n for away makes it wrap around properly with negative numbers