I dont know if my output is right!

[bits12345]

Problem:
Given the following function and it's results:

f(x,y) = x^y + y^x where x>0 and y>1

f(1,2) = 1^2 + 2^1 = 3
f(2,3) = 2^3 + 3^2 = 17
f(3,4) = 3^4 + 4^3 = 145
.
.
.
Using the above function denoted by f(x,y), and given that the initial values of x = 1 and y = 2, which increase by 1 each time, what are the last 4 digits of the sum of the first 15 function calls.


my code:

  #include <iostream>
#include <cmath>
#include <iomanip>
using namespace std;

main()
{
 	   double x = 1;
 	   double y = 2;
 	   double f, fx = 0;
 	   int counter;
 	   
 	               cout << setprecision(0);
 	   			   cout << fixed;
 	   		   	 
			 for (counter = 1; counter < 16; counter++)
	 	     {   
			  	 f = pow(x ,y) + pow (y, x);
			  	 cout << "\n" << counter <<". f(" << x << ", " << y << ") = " << x << "^" << y << " + " << y << "^" << x <<" = " << f;
			  	 fx = fx + f;
	x++;
	y++;		
			  	 			  	 		  	 			  			  	 			  	 
			 }		

	   cout <<" \n\n" << fx;			 	 
   	   cout << "\n\n";	
   	   
	   system("pause");
	   return 0; 	   
 	   
}

output:


1. f(1, 2) = 1^2 + 2^1 = 3
2. f(2, 3) = 2^3 + 3^2 = 17
3. f(3, 4) = 3^4 + 4^3 = 145
4. f(4, 5) = 4^5 + 5^4 = 1649
5. f(5, 6) = 5^6 + 6^5 = 23401
6. f(6, 7) = 6^7 + 7^6 = 397585
7. f(7, 8) = 7^8 + 8^7 = 7861953
8. f(8, 9) = 8^9 + 9^8 = 177264449
9. f(9, 10) = 9^10 + 10^9 = 4486784401
10. f(10, 11) = 10^11 + 11^10 = 125937424601
11. f(11, 12) = 11^12 + 12^11 = 3881436747409
12. f(12, 13) = 12^13 + 13^12 = 130291290501553
13. f(13, 14) = 13^14 + 14^13 = 4731091158953433
14. f(14, 15) = 14^15 + 15^14 = 184761021583202850
15. f(15, 16) = 15^16 + 16^15 = 7721329860319737900

sum = 7910956276398901200


when i type 1200 as the last 4 digits of the sum of the first 15 first calls it says im wrong. :(

HELP!

[bits12345]

bump

[condor]

Who said that you are wrong is right because the last 4 digits of the sum of the first 15 calls of that function are... 1049. bump!

[Chervil]

Looks like the type double is not precise enough here. (Not enough significant digits). You could try changing it to long double instead.

Failing that, use the unsigned long long type which is an integer, and you would have to write your own code to calculate the pow() function.

[condor]

Here is the output of my code:

  1. f(1, 2) = 1^2 + 2^1 = 3
  2. f(2, 3) = 2^3 + 3^2 = 17
  3. f(3, 4) = 3^4 + 4^3 = 145
  4. f(4, 5) = 4^5 + 5^4 = 1649
  5. f(5, 6) = 5^6 + 6^5 = 23401
  6. f(6, 7) = 6^7 + 7^6 = 397585
  7. f(7, 8) = 7^8 + 8^7 = 7861953
  8. f(8, 9) = 8^9 + 9^8 = 177264449
  9. f(9, 10) = 9^10 + 10^9 = 4486784401
 10. f(10, 11) = 10^11 + 11^10 = 125937424601
 11. f(11, 12) = 11^12 + 12^11 = 3881436747409
 12. f(12, 13) = 12^13 + 13^12 = 130291290501553
 13. f(13, 14) = 13^14 + 14^13 = 4731091158953433
 14. f(14, 15) = 14^15 + 15^14 = 184761021583202849
 15. f(15, 16) = 15^16 + 16^15 = 7721329860319737601

The sum of the above 15 values = 7910956276398901049

All values ​​were checked with Linux great precision arithmetic bc. A windows port of this bc arithmetic at:
http://gnuwin32.sourceforge.net/packages/bc.htm

[JLBorges]

> what are the last 4 digits ...

Use modular arithmetic.

#include <iostream>

constexpr int N = 10000 ;

long long pow( int a, int b ) // non-negative b
{
    long long result = 1 ;
    for( ; b > 0 ; --b ) result *= a ;
    return result ;
}

int pow_mod_n( int a, int b ) { return pow(a,b) % N ; }

int f_mod_n( int a, int b ) { return ( pow_mod_n(a,b) + pow_mod_n(b,a) ) % N ; }

int main()
{
     int sum = 0 ;
     for( int n = 1 ; n < 16 ; ++n ) sum += f_mod_n( n, n+1 ) ;
     std::cout << sum % N << '\n' ;
}

http://coliru.stacked-crooked.com/a/4627339a7db30120

[condor]

@bits12345

The code which has given the outputs in my above posts:

#include <iostream>
#include <iomanip>

long double myPow(int, int);

int main()
{
	int x = 1;
 	int y = 2;
	int counter;
	long double f;
	long double fx = 0.L;
 	std::cout << std::setprecision(0);
 	std::cout << std::fixed;
 	   		   	 
	for (counter = 1; counter < 16; ++counter)
	{   
		f = myPow(x, y) + myPow (y, x);
		std::cout << '\n' << std::setw(3) << counter <<". f(" << x << ", " << y << ") = " << x << "^" << y 
		<< " + " << y << "^" << x <<" = " << f;
		fx = fx + f;
		++x;
		++y;		
	}		

	std::cout << "\n\nThe sum of the above 15 values = " << fx;			 	 
        std::cout << '\n';	
   	   
        system("pause"); 
	return 0; 	   
}

long double myPow(int u, int v)
{
	long double f = 1.L;
	for(int i = 0; i < v; ++i)
		f *= u;
	return f;
}

Hope this helps.

[bits12345]

Thanks for the help you guys. I'll try to understand furthermore what you guys have posted. Haha