Keeping the order of operations while doing a sequential calculation - c++

I am spending my evening doing some programming problems from Kattis. There is one part of the problem 4 thought that I am stuck on.
Given a number, the program is supposed to return the operations (+, -, * or /) required between 4 fours to achieve that number.
For example, the input
9
would result in the output
4 + 4 + 4 / 4 = 9
My solution (not efficient, but simple) is to evaluate all possible ways to combine the operators above and see if any of the combinations achieve the wanted result.
To do this I have written the function seen below. It takes in an array of chars which are the operators to be evaluated (uo[3], could look like {+, /, *}), and the wanted result as an integer (expRes).
bool check(char uo[3], int expRes) {
int res = 4;
for(int oPos = 2; oPos >= 0; oPos--) {
switch (uo[oPos]) {
case '+' : res += 4; break;
case '-' : res -= 4; break;
case '*' : res *= 4; break;
case '/' : res /= 4; break;
}
}
return res == expRes;
}
I realized that this "sequential" approach comes with a problem: it doesn't follow the order of operations. If I was to call the function with
uo = {+, -, /}
and
expRes = 7 it would return false since 4 + 4 = 8, 8 - 4 = 4, 4 / 4 = 1.
The real answer is obviously true, since 4 + 4 - 4 / 4 = 7.
Can any of you think of a way to rewrite the function so that the evaluation follows the order of operations?
Thanks in advance!

Its an easy problem if you look at it.
You are restricted with four 4's and three operators in between, that is you already know your search space. So one solution is to generate the complete search space which is O(n^3) = 4^3 = 64 total equations, where n is the number of operators. Keep the answer to these solutions as a <key, value> pair so that look up to the input of test case is O(1).
Step wise you'd do.
Generate Complete Sequence and store them as key, value pairs
Take Input from test cases
Check if key exists, if yes print the sequence, else print that the sequence doesn't exist
Solution would take 64*1000 operations, which can easily be computed with in a second and would avoid Time Limit Exceeded Error that usually these competitions have
in Code form (most of it is incomplete):
// C++ Syntax
map<int, string> mp;
void generateAll() {
// generate all equations
}
void main () {
generateAll();
int n, t; scanf("%d", &t);
while (t--) {
scanf("%d", &n);
if ( mp.find(n) != mp.end() )
// equation exists to the input
else
// equation doesn't exist for the input
}
}

Related

pigeon hole / multiple numbers

input : integer ( i'll call it N ) and (1 <= N <= 5,000,000 )
output : integer, multiple of N and only contains 0,7
Ex.
Q1 input : 1 -> output : 7 ( 7 mod 1 == 0 )
Q2 input : 2 -> output : 70 ( 70 mod 2 == 0 )
#include <string>
#include <iostream>
using namespace std;
typedef long long ll;
int remaind(string num, ll m)
{
ll mod = 0;
for (int i = 0; i < num.size(); i++) {
int digit = num[i] - '0';
mod = mod * 10 + digit;
mod = mod % m;
}
return mod;
}
int main()
{
int n;
string ans;
cin >> n;
ans.append(n, '7');
for (int i = ans.length() - 1; i >= 0; i--)
{
if (remaind(ans, n) == 0)
{
cout << ans;
return 0;
}
ans.at(i) = '0';
}
return 0;
}
is there a way to lessen the time complexity?
i just tried very hard and it takes little bit more time to run while n is more than 1000000
ps. changed code
ps2. changed code again because of wrong code
ps3. optimize code again
ps4. rewrite post
Your approach is wrong, let's say you divide "70" by 5. Then you result will be 2 which is not right (just analyze your code to see why that happens).
You can really base your search upon numbers like 77777770000000, but think more about that - which numbers you need to add zeros and which numbers you do not.
Next, do not use strings! Think of reminder for a * b if you know reminder of a and reminder of b. When you program it, be careful with integer size, use 64 bit integers.
Now, what about a + b?
Finally, find reminders for numbers 10, 100, 1000, 10000, etc (once again, do not use strings and still try to find reminder for any power of 10).
Well, if you do all that, you'll be able to easily solve the whole problem.
May I recommend any of the boost::bignum integer classes?
I suspect uint1024_t (or whatever... they also have 128, 256, and 512, bit ints already typedefed, and you can declare your own easily enough) will meet your needs, allowing you to perform a single %, rather than one per iteration. This may outweigh the performance lost when using bignum vs c++'s built-in ints.
2^1024 ~= 1.8e+308. Enough to represent any 308 digit number. That's probably excessive.
2^512 ~= 1.34e+154. Good for any 154 digit number.
etc.
I suspect you should first write a loop that went through n = 4e+6 -> 5e+6 and wrote out which string got the longest, then size your uint*_t appropriately. If that longest string length is more than 308 characters, you could just whip up your own:
typedef number<cpp_int_backend<LENGTH, LENGTH, unsigned_magnitude, unchecked, void> > myReallyUnsignedBigInt;
The modulo operator is probably the most expensive operation in that inner loop. Performing once per iteration on the outer loop rather than at the inner loop (O(n) vs O(n^2)) should save you quite a bit of time.
Will that plus the whole "not going to and from strings" thing pay for bignum's overhead? You'll have to try it and see.

Cross sum calculation, Can anyone explain the code please?

i'm going to learn C++ at the very beginning and struggling with some challenges from university.
The task was to calculate the cross sum and to use modulo and divided operators only.
I have the solution below, but do not understand the mechanism..
Maybe anyone could provide some advice, or help to understand, whats going on.
I tried to figure out how the modulo operator works, and go through the code step by step, but still dont understand why theres need of the while statement.
#include <iostream>
using namespace std;
int main()
{
int input;
int crossSum = 0;
cout << "Number please: " << endl;
cin >> input;
while (input != 0)
{
crossSum = crossSum + input % 10;
input = input / 10;
}
cout << crossSum << endl;
system ("pause");
return 0;
}
Lets say my input number is 27. cross sum is 9
frist step: crossSum = crossSum + (input'27' % 10 ) // 0 + (modulo10 of 27 = 7) = 7
next step: input = input '27' / 10 // (27 / 10) = 2.7; Integer=2 ?
how to bring them together, and what does the while loop do? Thanks for help.
Just in case you're not sure:
The modulo operator, or %, divides the number to its left by the number to its right (its operands), and gives the remainder. As an example, 49 % 5 = 4.
Anyway,
The while loop takes a conditional statement, and will do the code in the following brackets over and over until that statement becomes false. In your code, while the input is not equal to zero, do some stuff.
To bring all of this together, every loop, you modulo your input by 10 - this will always return the last digit of a given Base-10 number. You add this onto a running sum (crossSum), and then divide the number by 10, basically moving the digits over by one space. The while loop makes sure that you do this until the number is done - for example, if the input is 104323959134, it has to loop 12 times until it's got all of the digits.
It seems that you are adding the digits present in the input number. Let's go through it with the help of an example, let input = 154.
Iteration1
crossSum= 0 + 154%10 = 4
Input = 154/10= 15
Iteration2
crossSum = 4 + 15%10 = 9
Input = 15/10 = 1
Iteration3
crossSum = 9 + 1%10 = 10
Input = 1/10 = 0
Now the while loop will not be executed since input = 0. Keep a habit of dry running through your code.
#include <iostream>
using namespace std;
int main()
{
int input;
int crossSum = 0;
cout << "Number please: " << endl;
cin >> input;
while (input != 0) // while your input is not 0
{
// means that when you have 123 and want to have the crosssum
// you first add 3 then 2 then 1
// mod 10 just gives you the most right digit
// example: 123 % 10 => 3
// 541 % 10 => 1 etc.
// crosssum means: crosssum(123) = 1 + 2 + 3
// so you need a mechanism to extract each digit
crossSum = crossSum + input % 10; // you add the LAST digit to your crosssum
// to make the number smaller (or move all digits one to the right)
// you divide it by 10 at some point the number will be 0 and the iteration
// will stop then.
input = input / 10;
}
cout << crossSum << endl;
system ("pause");
return 0;
}
but still dont understand why theres need of the while statement
Actually, there isn't need (in literal sense) for, number of digits being representable is limited.
Lets consider signed char instead of int: maximum number gets 127 then (8-bit char provided). So you could do:
crossSum = number % 10 + number / 10 % 10 + number / 100;
Same for int, but as that number is larger, you'd need 10 summands (32-bit int provided)... And: You'd always calculate the 10 summands, even for number 1, where actually all nine upper summands are equal to 0 anyway.
The while loop simplifies the matter: As long as there are yet digits left, the number is unequal to 0, so you continue, and as soon as no digits are left (number == 0), you stop iteration:
123 -> 12 -> 1 -> 0 // iteration stops, even if data type is able
^ ^ ^ // to store more digits
Marked digits form the summands for the cross sum.
Be aware that integer division always drops the decimal places, wheras modulo operation delivers the remainder, just as in your very first math lessons in school:
7 / 3 = 2, remainder 1
So % 10 will give you exactly the last (base 10) digit (the least significant one), and / 10 will drop this digit afterwards, to go on with next digit in next iteration.
You even could calculate the cross sum according to different bases (e. g. 16; base 2 would give you the number of 1-bits in binary representation).
Loop is used when we want to repeat some statements until a condition is true.
In your program, the following statements are repeated till the input becomes 0.
Retrieve the last digit of the input. (int digit = input % 10;)
Add the above retrieved digit to crosssum. (crosssum = crosssum + digit;)
Remove the last digit from the input. (input = input / 10;)
The above statements are repeated till the input becomes zero by repeatedly dividing it by 10. And all the digits in input are added to crosssum.
Hence, the variable crosssum is the sum of the digits of the variable input.

Multiply numbers which are divisible by 3 and less than 10 with a while loop in c++?

In C++, I should write a program where the app detects which numbers are divisible by 3 from 1 till 10 and then multiply all of them and print the result. That means that I should multiply 3,6,9 and print only the result, which is 162, but I should do it by using a "While" loop, not just multiplying the 3 numbers with each other. How should I write the code of this? I attached my attempt to code the problem below. Thanks
#include <iostream>
using namespace std;
int main() {
int x, r;
int l;
x = 1;
r = 0;
while (x < 10 && x%3==0) {
r = (3 * x) + 3;
cout << r;
}
cin >> l;
}
Firstly your checking the condition x%3 == 0 brings you out of your while - loop right in the first iteration where x is 1. You need to check the condition inside the loop.
Since you wish to store your answer in variable r you must initialize it to 1 since the product of anything with 0 would give you 0.
Another important thing is you need to increment the value of x at each iteration i.e. to check if each number in the range of 1 to 10 is divisible by 3 or not .
int main()
{
int x, r;
int l;
x = 1;
r = 1;
while (x < 10)
{
if(x%3 == 0)
r = r*x ;
x = x + 1; //incrementing the value of x
}
cout<<r;
}
Lastly I have no idea why you have written the last cin>>l statement . Omit it if not required.
Ok so here are a few hints that hopefully help you solving this:
Your approach with two variables (x and r) outside the loop is a good starting point for this.
Like I wrote in the comments you should use *= instead of your formula (I still don't understand how it is related to the problem)
Don't check if x is dividable by 3 inside the while-check because it would lead to an too early breaking of the loop
You can delete your l variable because it has no affect at the moment ;)
Your output should also happen outside the loop, else it is done everytime the loop runs (in your case this would be 10 times)
I hope I can help ;)
EDIT: Forget about No.4. I didn't saw your comment about the non-closing console.
int main()
{
int result = 1; // "result" is better than "r"
for (int x=1; x < 10; ++x)
{
if (x%3 == 0)
result = result * x;
}
cout << result;
}
or the loop in short with some additional knowledge:
for (int x=3; x < 10; x += 3) // i know that 3 is dividable
result *= x;
or, as it is c++, and for learning purposes, you could do:
vector<int> values; // a container holding integers that will get the multiples of 3
for (int x=1; x < 10; ++x) // as usual
if ( ! x%3 ) // same as x%3 == 0
values.push_back(x); // put the newly found number in the container
// now use a function that multiplies all numbers of the container (1 is start value)
result = std::accumulate(values.begin(), values.end(), 1, multiplies<int>());
// so much fun, also get the sum (0 is the start value, no function needed as add is standard)
int sum = std::accumulate(values.begin(), values.end(), 0);
It's important to remember the difference between = and ==. = sets something to a value while == compares something to a value. You're on the right track with incrementing x and using x as a condition to check your range of numbers. When writing code I usually try and write a "pseudocode" in English to organize my steps and get my logic down. It's also wise to consider using variables that tell you what they are as opposed to just random letters. Imagine if you were coding a game and you just had letters as variables; it would be impossible to remember what is what. When you are first learning to code this really helps a lot. So with that in mind:
/*
- While x is less than 10
- check value to see if it's mod 3
- if it's mod 3 add it to a sum
- if not's mod 3 bump a counter
- After my condition is met
- print to screen pause screen
*/
Now if we flesh out that pseudocode a little more we'll get a skeletal structure.
int main()
{
int x=1//value we'll use as a counter
int sum=0//value we'll use as a sum to print out at the end
while(x<10)//condition we'll check against
{
if (x mod 3 is zero)
{
sum=x*1;
increment x
}
else
{
increment x
}
}
//screen output the sum the sum
//system pause or cin.get() use whatever your teacher gave you.
I've given you a lot to work with here you should be able to figure out what you need from this. Computer Science and programming is hard and will require a lot of work. It's important to develop good coding habits and form now as it will help you in the future. Coding is a skill like welding; the more you do it the better you'll get. I often refer to it as the "Blue Collar Science" because it's really a skillset and not just raw knowledge. It's not like studying history or Biology (minus Biology labs) because those require you to learn things and loosely apply them whereas programming requires you to actually build something. It's like welding or plumbing in my opinion.
Additionally when you come to sites like these try and read up how things should be posted and try and seek the "logic" behind the answer and come up with it on your own as opposed to asking for the answer. People will be more inclined to help you if they think you're working for something instead of asking for a handout (not saying you are, just some advice). Additionally take the attitude these guys give you with a grain of salt, Computer Scientists aren't known to be the worlds most personable people. =) Good luck.

Efficient Exponentiation For HUGE Numbers (I'm Talking Googols)

I am in the midst of solving a simple combination problem whose solution is 2^(n-1).
The only problem is 1 <= n <= 2^31 -1 (max value for signed 32 bit integer)
I tried using Java's BigInteger class but It times out for numbers 2^31/10^4 and greater, so that clearly doesn't work out.
Furthermore, I am limited to using only built-in classes for Java or C++.
Knowing I require speed, I chose to build a class in C++ which does arithmetic on strings.
Now, when I do multiplication, my program multiplies similarly to how we multiply on paper for efficiency (as opposed to repeatedly adding the strings).
But even with that in place, I can't multiply 2 by itself 2^31 - 1 times, it is just not efficient enough.
So I started reading texts on the problem and I came to the solution of...
2^n = 2^(n/2) * 2^(n/2) * 2^(n%2) (where / denotes integer division and % denotes modulus)
This means I can solve exponentiation in a logarithmic number of multiplications. But to me, I can't get around how to apply this method to my code? How do I choose a lower bound and what is the most efficient way to keep track of the various numbers that I need for my final multiplication?
If anyone has any knowledge on how to solve this problem, please elaborate (example code is appreciated).
UPDATE
Thanks to everyone for all your help! Clearly this problem is meant to be solved in a realistic way, but I did manage to outperform java.math.BigInteger with a power function that only performs ceil(log2(n)) iterations.
If anyone is interested in the code I've produced, here it is...
using namespace std;
bool m_greater_or_equal (string & a, string & b){ //is a greater than or equal to b?
if (a.length()!=b.length()){
return a.length()>b.length();
}
for (int i = 0;i<a.length();i++){
if (a[i]!=b[i]){
return a[i]>b[i];
}
}
return true;
}
string add (string& a, string& b){
if (!m_greater_or_equal(a,b)) return add(b,a);
string x = string(a.rbegin(),a.rend());
string y = string(b.rbegin(),b.rend());
string result = "";
for (int i = 0;i<x.length()-y.length()+1;i++){
y.push_back('0');
}
int carry = 0;
for (int i =0;i<x.length();i++){
char c = x[i]+y[i]+carry-'0'-'0';
carry = c/10;
c%=10;
result.push_back(c+'0');
}
if (carry==1) result.push_back('1');
return string(result.rbegin(),result.rend());
}
string multiply (string&a, string&b){
string row = b, tmp;
string result = "0";
for (int i = a.length()-1;i>=0;i--){
for (int j= 0;j<(a[i]-'0');j++){
tmp = add(result,row);
result = tmp;
}
row.push_back('0');
}
return result;
}
int counter = 0;
string m_pow (string&a, int exp){
counter++;
if(exp==1){
return a;
}
if (exp==0){
return "1";
}
string p = m_pow(a,exp/2);
string res;
if (exp%2==0){
res = "1"; //a^exp%2 is a^0 = 1
} else {
res = a; //a^exp%2 is a^1 = a
}
string x = multiply(p,p);
return multiply(x,res);
//return multiply(multiply(p,p),res); Doesn't work because multiply(p,p) is not const
}
int main(){
string x ="2";
cout<<m_pow(x,5000)<<endl<<endl;
cout<<counter<<endl;
return 0;
}
As mentioned by #Oli's answer, this is not a question of computing 2^n as that's trivially just a 1 followed by 0s in binary.
But since you want to print them out in decimal, this becomes a question of how to convert from binary to decimal for very large numbers.
My answer to that is that it's not realistic. (I hope this question just stems from curiosity.)
You mention trying to compute 2^(2^31 - 1) and printing that out in decimal. That number is 646,456,993 digits long.
Java BigInteger can't do it. It's meant for small numbers and uses O(n^2) algorithms.
As mentioned in the comments, there are no built-in BigNum libraries in C++.
Even Mathematica can't handle it: General::ovfl : Overflow occurred in computation.
Your best bet is to use the GMP library.
If you're just interested in seeing part of the answer:
2^(2^31 - 1) = 2^2147483647 =
880806525841981676603746574895920 ... 7925005662562914027527972323328
(total: 646,456,993 digits)
This was done using a close-sourced library and took roughly 37 seconds and 3.2 GB of memory on a Core i7 2600K # 4.4 GHz including the time needed to write all 646 million digits to a massive text file.
(It took notepad longer to open the file than needed to compute it.)
Now to answer your question of how to actually compute such a power in the general case, #dasblinkenlight has the answer to that which is a variant of Exponentiation by Squaring.
Converting from binary to decimal for large numbers is a much harder task. The standard algorithm here is Divide-and-Conquer conversion.
I do not recommend you try to implement the latter - as it's far beyond the scope of starting programmers. (and is also somewhat math-intensive)
You don't need to do any multiplication at all. 2^(n-1) is just 1 << (n-1), i.e. 1 followed by (n-1) zeros (in binary).
The easiest way to apply this method in your code is to apply it the most direct way - recursively. It works for any number a, not only for 2, so I wrote code that takes a as a parameter to make it more interesting:
MyBigInt pow(MyBigInt a, int p) {
if (!p) return MyBigInt.One;
MyBigInt halfPower = pow(a, p/2);
MyBigInt res = (p%2 == 0) ? MyBigInt.One : a;
return res * halfPower * halfPower;
}

first and last k digits of number n^n

i have written a c++ code for generating first and last k digits of a number as large as 10^9. (k<=9).
cin>>n>>k;
cout << (unsigned long)floor(pow(10.0, modf(n*log10((double)n), &dummy) + k - 1)) << " "; // code that prints the first k digits
long long int ans = foo(n,k); // function that prints the last k digits
if(ans==0)
{
for(int i=0;i<k;i++) cout << "0";
}
else{
stringstream ss;
string s;
ss<<ans;
ss>>s;
if(s.size()!=k)
{
for(int i=0;i<(k-s.size());i++)
s="0"+s;
}
cout<<s;
}
where function foo() is:
long long int foo(int n, int k) // code of the function
{
long long int m=1;
for(; k > 0; k--) m*=10;
long long int r=1, t=n % m;
while(n)
{
if (n % 2)
r = r * t % m;
t = t * t % m;
n >>= 1;
}
return r;
}
this gives me output as:
if given 9 and 3 as inputs, it gives first and last 3 digits of 9 to the power 9 (9^9) i.e. 387 and 489. But I m still missing out some test cases.
Can anyone please help me finding out the test case for which my code wouldn't work ?
1 ≤ n ≤ 109, 1 ≤ k ≤ 9
the problem statement: http://www.codechef.com/problems/MARCHA4/
If n^n <= 10^9, in which case your code seems to work fine. However, if you allow bigger n, say 11^11, and ask for the last 4 digits of that, which are 0611, your code will only print 611. Basically, it doesn't print any leading zeroes when it should.
This doesn't really answer the question, and its almost trivially easy, but I figure it might be worth sharing. If there were a "long comment" capability I'd be using it.
EDIT: just noticed using str instead of repr will eliminate the L on its own
def firstAndLastDig(k, num):
s = str(num)
return (s[:k], s[-k:])
def firstAndLastDigSelfExp(k,n):
return firstAndLastDig(k,n**n)
Overflow is not an issue (the only thing is dealing with the L if you use repr instead of str),
firstAndLastDigSelfExp(6,12)
('891610', '448256')
firstAndLastDigSelfExp(42,491)
('209417336844579728122309696211520194012462', '160453713040914743773217804810667483135091')
And neither are leading zeroes
>>> firstAndLastDigSelfExp(4,9)
('3874', '0489')
This isn't do say the modular logs and stuff aren't cool - on the contrary I really liked reading about how you did this without generating the entire number. I didn't know about modf at all until reading OP's question and the body of foo is very interesting.
I think the problem is using floating point. Finding the first digit of a number actually requires perfect precision.
Unfortunately, the contest judge evidently doesn't understand that "number of significant digits" != "number of correct digits".
Perhaps there is some clever way to exactly compute (n*n, n = 10*9) without exhausting memory, but finding the first digits of a very good estimate is simply not the same as finding the first digits of the answer.
Assume that k = 9. Now, m = 1e9, and t <= 1e9 - 1.
t * t then may be as high as 1e18 - 2e9 + 1, which needs ... 59.8 bits.
Ok, not a problem with a 64-bit long long int, which has 63 bits of magnitude (and 1 of sign), but I'll leave this here so others don't repeat the same analysis.
Are you told that n is a positive integer? For example, (-8)^(-8) is perfectly well expressible in decimal but your program can't handle it.