I use a quick_pow func to calculate a^^b (a=2, b is more than 400), and at the same time, I also need mod to get lower positions num as the output of leetcode exercise.
With hours' checking, I finally located the problem coming from my qpow() func. But after I change it to an easier way in qpow2(), it does give correct answer. So I tried a simple code to see the difference. As you can see in my screenshot, qpow() gets wrong in "-83648" and then keeps "0" after that. Could you please let me know what is going on here? Thanks a lot.
#include<iostream>
#include<stdlib.h>
#include<cstring>
#include<algorithm>
#include<math.h>
using namespace std;
int qpow(int n) {
int x = 1, a = 2;
while (n > 0) {
if (n % 2) {
x *= a;
x %= 100000;
};
a *= a;
a %= 100000;
n /= 2;
};
return x;
};
int qpow2(int n) {
int x = 1, a = 2;
while (n--)
{
x *= a;
x %= 100000;
};
return x;
};
int main() {
int n;
cin >> n;
for (int i = 10; i < n; i++)
{
cout << qpow(i) << " | ";
cout << qpow2(i) << " | ";
cout << INT_MAX << endl;
};
return 0;
};
TRYING 2^^50
Related
My code:
#include <iostream>
#include <bits/stdc++.h>
using namespace std;
int BinaryToDecimal(int n)
{
int ans = 0;
int x = 1;
while (n > 0)
{
int y = n % 10;
ans = ans + x * y;
x = x * 2;
n = n / 10;
}
return ans;
}
int DecimalToBinary(int num)
{
vector<int> vect;
while (num > 0)
{
vect.push_back(num % 2);
num = num / 2;
}
int s = vect.size();
int i = s - 1;
for (i = s - 1; i >= 0; i--)
{
cout << vect.at(i);
}
return vect.at(i);
}
int main()
{
int a, b;
cout << "Enter first number: " << endl;
cin >> a;
cout << "Enter second number: " << endl;
cin >> b;
int a_deci = BinaryToDecimal(a);
int b_deci = BinaryToDecimal(b);
int sum = a_deci + b_deci;
DecimalToBinary(sum);
cout << endl;
return 0;
}
Output:
Enter first number:
10101
Enter second number:
11010
101111terminate called after throwing an instance of 'std::out_of_range'what(): vector::_M_range_check: __n (which is 18446744073709551615) >= this->size() (which is 6)
What does this error message mean and how do I fix it?
After this for loop
for (i = s - 1; i >= 0; i--)
{
cout << vect.at(i);
}
the variable i is equal to -1.
So the next call of the member function at with the value equal to -1 (that yields a very big number of the unsigned type std::vector<int>::size_type)
return vect.at(i);
throws the exception.
It seems you need to return from the function the whole vector elements of which will represent a number in the binary form.
Instead of the container std::vector<int> it will be better to use std::bitset.
This is what I came up with
#include <iostream>
using namespace std;
int serialNumber = 1;
Would recursion be better?
int factorial(int n)
{
int k=1;
for(int i=1;i<=n;++i)
{
k=k*i;
}
return k;
}
How can I go about doing this in a single for loop?
Or is this the best way?
int main()
{
int a;
int b;
int c;
int fact1;
int fact2;
int fact3;
for (a=1;a < 11;a++)
{
fact1 = factorial(a);
for (b=1;b < 11;b++)
{
fact2 = factorial(b);
for (c=1;c < 11;c++)
{
fact3 = factorial(c);
cout << serialNumber << " : ";
int LHS = fact1 + fact2 + fact3;
if (LHS == a * b * c)
{
cout << "Pass:" <<" "<< a << " & " << b << " & " << c << endl;
}
else
{
cout << "Fail" <<endl;
}
serialNumber++;
}
c = 1;
}
b = 1;
}
return 0;
}
I am being forced to add more none code into it.
Thanks for the help!
Don't know if this is helps,but>
check for minimum of A,B,C
A!+B!+C! = (min(A,B,C)!)*(1+((min+1..restfact1)!)+((min+1..restfact2)!))
So, you can calculate the minimum factorial and than re-use it for calculating others.
On the other hand, you can calculate only the maximum factorial and store its results in the array, and re-use pre-calculated values for finding factorial of smaller numbers
Other implication is that the minimum number can be reduced
restfact1 * restfact2 = ((min-1)!)*(1+((min+1..restfact1)!)+((min+1..restfact2)!))
Part of the question was how can this be done in a single loop and this is one way to do that.
I don't think this is a better way of doing it, but the question was asked:
constexpr int bound = 10;
int Factorials[bound + 1];
for (int i = 1; i <= bound; ++i) Factorials[i] = Factorial(i);
for (int i = 0; i < bound * bound * bound; ++i) {
int s = i + 1;
int a = i;
int c = 1 + a % bound;
a /= bound;
int b = 1 + a % bound;
a /= bound;
++a;
cout << s << " : ";
int LHS = Factorials[a] + Factorials[b] + Factorials[c];
if (LHS == a * b * c)
...
}
I made a simple recursion program for this question http://www.spoj.com/problems/COINS/, but whenever recursion happens my class variables lose their value and store the value from the recursion loop.
Here's the code:
#include<iostream>
using namespace std;
class a
{
public:
int c = 0, d = 0, b = 0, x = 0;
int recur(int n)
{
b = (n / 2);
if (b >= 12)
{
b = recur(b);
}
c = (n / 3);
if (c >= 12)
{
c = recur(c);
}
d = (n / 4);
if (d >= 12)
{
d = recur(d);
}
x = b + c + d;
return x;
}
};
int main()
{
int n;
while(cin)
{
cin >> n;
int b = 0, r = 0;
a abc;
r = (n > abc.recur(n)) ? (n) : (abc.recur(n));
cout << r << endl;
}
return 0;
}
So for input 12, I'll be getting 13 but for the input value of 44 I'm getting 44.
This could be a working solution:
#include <iostream>
using namespace std;
int changeToDollars(int bytelandians) {
int byTwo = bytelandians / 2;
int byThree = bytelandians / 3;
int byFour = bytelandians / 4;
int sum = byTwo + byThree + byFour;
if (sum < bytelandians) {
return bytelandians;
} else {
return changeToDollars(byTwo) + changeToDollars(byThree) + changeToDollars(byFour);
}
}
int main() {
int bytelandians;
cout << "How much bytelandians?: ";
while (cin >> bytelandians) {
cout << "Corresponding $: " << changeToDollars(bytelandians) << endl;
cout << "How much bytelandians?: ";
}
return 0;
}
The changeToDollars function, using a simple recursive algorithm, exchanges each single Byteland coin into the corresponding three ones with minor value, until the overall converted amount is advantageous.
I am using this New and improved code I corrected in order to solve this question I have.
I am using modular Exponentiation to use the formula [a^k mod n] to get my answer for an assignment I had to do where I was required to code it in two steps.
First int k must be converted to a binary
representation K consisting of a list of 0s and 1s. Second, Modular Exponentiation must be performed
using a, n and K[] as arguments..
Earlier My code was incorrect and was able to correct it.
The Problem I now face is that when I google the online calculator for modular Exponentiation of 5^3 % 13, it should == 8
The result that I get from my code is 5.
I am trying to understand if there something minor I'm missing from the code or my math is wrong? Thanks
#include <iostream>
#include <vector>
using namespace std;
vector <int> BinaryK(int k);
int ModularExpo(int a, vector <int> & k, int n);
int main()
{
int a = 0;
int k = 0;
int n = 0;
cout << "a^k % n" << endl;
cout << "a = ";
cin >> a;
cout << "k = ";
cin >> k;
cout << "n = ";
cin >> n;
vector<int> B = BinaryK(k);
int result = ModularExpo(a, B, n);
cout << "a ^ k mod n == " << result << endl;
return 0;
}
// c == b^e % m
vector<int> BinaryK(int k)
{
vector<int> K; //hint: make K a vector
int tmp = k;
while (tmp > 0)
{
K.push_back(tmp % 2); //hint: use pushback
tmp = tmp / 2;
}
return K;
}
int ModularExpo(int a, vector<int> & K, int n)
{
if (n == 1)
return 0;
int b = 1;
if (K.size() == 0)
return b;
int A = a;
if (K[0] == 1)
b = a;
for (int i = 1; i < K.size() - 1; i++)
{
A = A * A % n;
if (K[i] == 1)
b = A*b % n;
}
return (b);
}
Change this one line:
for (int i = 1; i < K.size(); i++) // K.size() not K.size()-1
I'm trying to solve the 3n+1 problem using VS2010 c++,in small inputs it works well,but when it reaches 113383 it overflows.
Here is the problem link.
This is the code I'm using to solve this problem :
#include <iostream>
using namespace std;
int main(void) {
while (!cin.eof()) {
int i, j, maxCycle = 0, tmaxCycle = 0;
cin >> i >> j;
for (int x = i; x <= j; x++) {
int n = x;
tmaxCycle = 0;
while (n != 1) {
if ((float)(n/2) != (n/2.0)) {
n = 3*n + 1;
}
else {
n /= 2;
}
tmaxCycle += 1;
if (n < 0) {
int blah = 0; //just for the breakpoint
}
}
tmaxCycle += 1;
if (tmaxCycle > maxCycle) {
maxCycle = tmaxCycle;
}
}
cout << i << "\t" << j << "\t" << maxCycle << endl;
}
system("pause");
}
I made a breakpoint at line 15,and in this point values overflows
n=-1812855948
Use 64-bit unsigned integers. If those overflow, use a bignum library like the GNU Multiple Precision Library. Bignums give you unlimited precision and size.
This
if((float)(n/2)!=(n/2.0))
produces incorrect results, long before int overflows. Change it to
if ( n & 1)