INOI 2015 - Periodic Strings - c++

Question

Question: https://www.codechef.com/INOIPRAC/problems/INOI1502
Here's what I'd thought off -
Have a function, f(n) which finds the factors of n
If a factor, i, is found, call f(i)
for each value of n, the function also calculates the number of non periodic strings would be equal to 2^n - (the value returned by each of the function calls)
return the number of non periodic strings and store this number in an array to prevent
Then I just call the function, f(n) modulo n to get the output
It works for smaller values, but not for larger ones
For example, when n=35 & m=99999989
My code as of now:
#include <iostream>
#include <cmath>
using namespace std;
int arr[150100];
int ans[150100];
int check(int n){
if(arr[n]>0){
return arr[n];
}
else if(n == 1){
arr[n] = 2;
return 2;
}
if(n==2){
arr[n] = 2;
return 2;
}
for(int i =1 ;i<(n/2) +1;i++){
if(n%i == 0){
ans[n] -= check(i);//2+
}
}
arr[n] = ans[n];
return ans[n];
}
int main() {
int n,m;
cin>>n>>m;
for(int i=0;i<=150100;i++){
arr[i] = 0;
ans[i] = pow (2,i);
}
std::cout<<( check(n) )%m<<endl;
}
Full problem statement:
A string is any nonempty sequence of 0s and 1s. Examples of strings are 00, 101, 111000, 1, 0, 01. The length of a string is the number of symbols in it. For example, the length of 111000 is 6. If u and v are strings, then uv is the string obtained by concatenating u and v. For example if u = 110 and v = 0010 then uv = 1100010.
A string w is periodic if there exists a string v such that w = vn = vv · · · v (n times), for some n ≥ 2. Note that in this case the length of v is strictly less than that of w. For example, 110110 is periodic, because it is vv for v = 110.
Given a positive integer N , find the number of strings of length N which are not periodic. Report the answer modulo M. The non-periodic strings of length 2 are 10 and 01. The non- periodic strings of length 3 are 001, 010, 011, 100, 101, and 110.
Input format
A single line, with two space-separated integers, N and M.

Answer 1

Ok, I'll start from built in type you have choose, it's not the best choice for your example: n=35 & m=99999989. Generally size of int is 32 bits, so it capable to hold maximum 2^32. So for your example you should choose a type that capable to hold minimum 35 bits.
Long long is also not good choice since you use modulo function which applies on integers, if you want to apply modulo on type bigger than int, you will prefer to use function fmod, please see http://www.cplusplus.com/reference/cmath/fmod/.
In your implementation I would prefer to use double type, on most systems it's size is 64 bits, below is code with some corrections:
#include <iostream>
#include <cmath>
using namespace std;
double arr[150100];
double ans[150100];
double check(int n){
if(arr[n]>0){
return arr[n];
}
else if(n == 1){
arr[n] = 2;
return 2;
}
if(n==2){
arr[n] = 2;
return 2;
}
for(int i =1 ;i<(n/2) +1;i++){
if(n%i == 0){
ans[n] -= check(i);//2+
}
}
arr[n] = ans[n];
return ans[n];
}
int main() {
int n,m;
cin>>n>>m;
for(int i=0;i < 150100;i++){
arr[i] = 0;
ans[i] = pow(2.0,i);
}
std::cout<<static_cast<int>(fmod(check(n),m))<<endl;
}
Please note that this fix will work only for N up to 64, because on most systems double size is 64 bits.
The second issue that you should take into account is your "ans" array, you try to initialize it with values that are much more bigger than int or double capable to hold, values that are bigger than 2^64. in this case there will be truncated data in "ans".
For this task I would prefer another approach which includes modular exponentiation rules: ab mod m = (a mod m)(b mod m) mod m = (a(b mod m)) mod m
According to description in question 2 ≤ M ≤ 10^8, so it's enough to hold array of integers in this task.
For example, to calculate 2^150000 mod 10^8, instead of evaluating 2^150000 directly, do it step by step and take modulo at each step.