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The below code is to calculate 2^n where n is equal to 1 <= n <= 10^5. So to calculate such large numbers I have used concept of modular exponentian. The code is giving correct output but due to large number of test cases it is exceeding the time limit. I am not getting a way to minimize the solution so it consumes less time. As the "algo" function is called as many times as the number of test cases. So I want to put the logic used in "algo" function in the main() function so it consumes time less than 1 sec and also gives the correct output. Here "t" represents number of test cases and it's value is 1 <= t <= 10^5.

Any suggestions from your side would be of great help!!

#include<iostream>
#include<math.h>

using namespace std;

int algo(int x, int y){
  long m = 1000000007;
  if(y == 0){
    return 1;
  }

  int k = algo(x,y/2);

  if (y % 2 == 1){
    return ((((1ll * k * k) % m) * x) % m);
  } else if (y % 2 == 0){
    return ((1ll * k * k) % m);
  }
  
}

int main(void)
{
    int n, t, k;
    cin>>t; //t = number of test cases
    for ( k = 0; k < t; k++)
    {
        cin >> n;  //power of 2 

        cout<<"the value after algo is: "<<algo(2,n)<<endl;

    }
    return 0;
}
Lars Nielsen
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aditya_6494
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  • @UlrichEckhardt - my thoughts exactly. That's why I just ran it through a pretty-printer. – selbie May 10 '21 at 07:48
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    10^5 isn't that big, have you considered memoization or even precalculation of all the possible results? – Bob__ May 10 '21 at 08:02
  • Use `'\n'` instead of `endl`. The latter causes a flush, which might slow down anything consuming the data (especially if the data is being written to a file on a slow disk or over a network). It seems a bit strange to use recursion to solve this problem, too. Stylistically, you should not use `else if (y % 2 == 0)` -- instead, just use `else`, or simply nothing at all. – paddy May 10 '21 at 08:02
  • (a ⋅ b) mod m = [(a mod m) ⋅ (b mod m)] mod m (from wikipedia) you can surely leverage this identity to do much better than that. https://en.wikipedia.org/wiki/Modular_exponentiation – Alessandro Teruzzi May 10 '21 at 08:06

3 Answers3

0

You can make use of binary shifts to find powers of two

#include <iostream>
using namespace std;
int main()
{
    unsigned long long u = 1, w = 2, n = 10, p = 1000000007, r;
    //n -> power of two 
    while (n != 0)
    {
        if ((n & 0x1) != 0)
            u = (u * w) % p;

        if ((n >>= 1) != 0)
            w = (w * w) % p;
    }
    r = (unsigned long)u;
    cout << r;
    return 0;
}
Tharun K
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0

This is the function that I often use to calculate
Any integer X raised to power Y modulo M

C++ Function to calculate (X^Y) mod M

int power(int x, int y, const int mod = 1e9+7)
{
    int result = 1;
    x = x % mod;
    if (x == 0)
        return 0;
    while (y > 0)
    {
        if (y & 1)
            result = ( (result % mod) * (x % mod) ) % mod;
        y = y >> 1; // y = y / 2
        x = ( (x % mod) * (x % mod) ) % mod;
    }
    return result;
}
  • Remove the Mod if you don't want.
  • Time Complexity of this Function is O(log2(Y))
  • There can be a case of over flow so use int , long , long long etc as per your need.
Mr.HITMAN
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-1

Well your variables won't sustain the boundary test cases, introducing 2^10000, 1 <= n <= 10^5. RIP algorithms

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Fear not my friend, someone did tried to solve the problem https://www.quora.com/What-is-2-raised-to-the-power-of-50-000, you are looking for Piyush Michael's answer , here is his sample code

#include <stdio.h> 
int main() 
{ 
 int ul=16,000; 
 int rs=50,000; 
 int s=0,carry[ul],i,j,k,ar[ul]; 
 ar[0]=2; 
 for(i=1;i<ul;i++)ar[i]=0; 
 for(j=1;j<rs;j++) 
 {for(k=0;k<ul;k++)carry[k]=0; 
  for(i=0;i<ul;i++) 
  {ar[i]=ar[i]*2+carry[i]; 
   if(ar[i]>9) 
   {carry[i+1]=ar[i]/10; 
    ar[i]=ar[i]%10; 
   } 
  } 
 } 
 for(j=ul-1;j>=0;j--)printf("%d",ar[j]); 
 for(i=0;i<ul-1;i++)s+=ar[i]; 
 printf("\n\n%d",s); 
}
saumitra mallick
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