Task on the number of iterations - c++

Question

There is a number N
every iteration it becomes equal to (N*2)-1
I need to find out how many steps the number will be a multiple of the original N;
( 1≤ N ≤ 2 · 10 9 )
For example:
N = 7; count = 0
N_ = 7*2-1 = 13; count = 1; N_ % N != 0
N_ = 13*2-1 = 25; count = 2; N_ % N != 0
N_ = 25*2-1 = 49; count = 3; N_ % N == 0
Answer is 3
if it is impossible to decompose in this way, then output -1
#include <iostream>
using namespace std;
int main(){
int N,M,c;
cin >> N;
if (N%2==0) {
cout << -1;
return 0;
}
M = N*2-1;
c = 1;
while (M%N!=0){
c+=1;
M=M*2-1;
}
cout << c;
return 0;
}
It does not fit during (1 second limit). How to optimize the algorithm?
P.S All the answers indicated are optimized, but they don’t fit in 1 second, because you need to change the algorithm in principle. The solution was to use Euler's theorem.

Answer 1

The problem, as other answers have suggested, is equivalent to finding c such that pow(2, c) = 1 mod N. This is impossible if N is even, and possible otherwise (as your code suggests you know).
A linear-time approach is:
int c = 1;
uint64_t m = 2;
while (m != 1){
c += 1;
m = (2*m)%N;
}
printf("%d\n", c);
To solve this in 1 second, I don't think you can use a linear-time algorithm. The worst cases will be when N is prime and large. For example 1999999817 for which the above code runs in around 10 seconds on my laptop.
Instead, factor N into its prime factors. Solve 2^c = 1 mod p^k for each prime factor (where p^k appears in the prime factorization of N. Then combine the results using the Chinese Remainder theorem.
When finding the c for a given prime power, if k=1, the solution is c=p-1. When k is larger, the details are quite messy, but you can find a written solution here: https://math.stackexchange.com/questions/1863037/discrete-logarithm-modulo-powers-of-a-small-prime

Answer 2

The problem is that you're overflowing, the int data type only has 32 bits, and overflows 2^31-1 , in this problem you don't need to keep the actual value of M, you can just keep the modulo of n.
while (M%N!=0){
c+=1;
M=M*2-1;
M%=N
}
Edit:In addition, you don't actually need more than N iterations to check if a 0 mod exists, as there are only N different mods to N and it just keeps cycling. so you also need to keep that in mind in case there is no 0 mod.

Answer 3

There is no doubt that the main problem with your code is signed integer overflow
I added a print of M whenever M was changed (i.e. cout << M << endl;) and gave it the input 29. This is what I got:
57
113
225
449
897
1793
3585
7169
14337
28673
57345
114689
229377
458753
917505
1835009
3670017
7340033
14680065
29360129
58720257
117440513
234881025
469762049
939524097
1879048193
-536870911
-1073741823
-2147483647
1
1
1
1
... endless loop
As you see you have signed integer overflow. That is undefined behavior in C so anything may happen!! On my machine I ended up with a nasty endless loop. That must be fixed before considering performance.
The simple fix is to add a line like
M = M % N;
whenever M is changed. See the answer from #Malek
Besides that you shall also use an unsigned integer, i.e. use uint32_t for all variables.
However, that will not improve performance.
If you still have performance issue after the above fix, you can try this instead:
uint32_t N;
cin >> N;
if (N%2==0) {
cout << -1;
return 0;
}
// Alternative algorithm
uint32_t R,c;
R = 1;
c = 1;
while (R != N){
R = 2*R + 1;
if (R > N) R = R - N;
++c;
}
cout << c;
On my laptop this algorithm is 2.5 times faster when testing on all odd numbers in the range 1..100000. However, it might not be sufficient for all numbers in the range 1..2*10^9.
Also notice the use of uint32_t to avoid integer overflow.