Code this 1 2^2 + 3^3 4^4 + 5^5 -

Your task is to write a C++ program...

No, it is your task. We do not solve other peoples homework. Show us your work so far or ask specific questions, and we'll help, but nobody here will write your homework.

but nobody here will write your homework.
Did I do something wrong? (Usually I don't do this...) Well, you are free to remove it if you want.

Hey guys is this right? But when i compiled it, it gave me an error message..."Line 22; parse error at end of input" i don't see anything wrong!

#include <iostream>
using namespace std;

int main()
{
int sign = 1;
int sum = 0;
int n;

cout << "Enter a value: ";
cin >> n;

for(double i=1; i<=n; i++)
{
sum += sign * (int)pow(i, i);
sign *= -1;

cout << sum; //is this right? or should it be sign?

return 0;
}

Thanks Matt
By the way i'm new to this forum so i still havn't discovered where the quote things is...sorry : (

But Matt when i use int it gives me an error

"Line 16 : call of overloaded 'pow(int&, int&)' is ambiguous" but when i use double it's fine...

But i've just read the question again it told me to use long int.. I'm kinda caught out in the middle.

I'm using Quincy 2002 as the program and mingw as a compiler i think..
So can i do the same thing for long int?

Quote
for(int i=1; i<=n; i++)
{
sum += sign * (int)pow((double)i, (double)i);
sign *= -1;
}
Quote

Code
for(int i=1; i<=n; i++)
{
sum += sign * (int)pow((long int)i, (long int)i);
sign *= -1;
}
Code

oh oo... it seems like long int doesn't work either... bugger. Stupid compiler

Thanks T.G. and one more thing it asks me to "display the above series" as an output, do you think there is a looping way to do it or do i have to hardcode it?

I think it is possible to combine this printing into the loop that computes.

What about this solution?

template <int I, int J>
struct p
{
static const int ret = I* p<I, J-1>::ret;
};

template <int I>
struct p<I, 0>
{
static const int ret = 1;
};

template<int I>
struct q
{
static const int ret = q<I-1>::ret + ((I%2)?1:-1)*p<I, I>::ret;
};

template <>
struct q<1>
{
static const int ret = 1;
};

int main()
{
int j = q<6>::ret;
return 0;
}
:cool:

Somebody's been reading Alexandrescu, haven't they? :rolleyes:

In comp.lang.c++, this was the usual trick of "doing someone's homework", and that is to give a solution only an experienced C++ programmer would come up with. This way, if the student were to hand in the program, they will get questioned by the teacher, causing all sorts of problems for the student (and hopefully a lesson would be learned).

Regards,

Paul McKenzie

I like it... If we do everyone's homework for them, we'll eventually have to do their job for them when they get to that point...

In comp.lang.c++, this was the usual trick of "doing someone's homework", and that is to give a solution only an experienced C++ programmer would come up with. This way, if the student were to hand in the program, they will get questioned by the teacher, causing all sorts of problems for the student (and hopefully a lesson would be learned).

Isn't that mean? ;)

Isn't that mean? ;)

No, it's a good lesson. Not to speak of that showing of every now and then is good to your health :D

#include <stdlib.h>
#include <stdio.h>

int calcSeries(int N)
{
int k1, k2, n = (N+1)/2;

if (N<1) exit(-1);

if (N%2)
{
k1 = k2 = 1;
}
else
{
k1 = -1;
k2 = 3;
}

return k1*n*(k2+2*(n-1));
}

int main()
{
for (int i=1; i<20; ++i)
printf("%d \t%d\n", i, calcSeries(i));
return 0;
}

Guys can you check this for me?

#include <iostream>
using namespace std;

int main()
{
int sign = 1;
int sum = 0;
int n;
int i;

cout << "Enter a value: ";
cin >> n;

if((n <= 6) && (n > 0))
{
for(double i=1; i<=n; i++)
{
sum += sign * (int)pow((double)i, (double)i);
sign *= -1;
}
}
else
{
cout << "Invalid\n";
}

cout << "1";
for (i=2, i<=n, i++)
{
if mod(i%2)==0
{
cout << "-";
}
else
cout << "+";

cout << i << "^" << i;
}
cout << "=" << answer << endl;

return 0;
}

#include <stdlib.h>
#include <stdio.h>

int calcSeries(int N)
{
int k1, k2, n = (N+1)/2;

if (N<1) exit(-1);

if (N%2)
{
k1 = k2 = 1;
}
else
{
k1 = -1;
k2 = 3;
}

return k1*n*(k2+2*(n-1));
}

int main()
{
for (int i=1; i<20; ++i)
printf("%d \t%d\n", i, calcSeries(i));
return 0;
}

I will not try to understand what you did here! Did you and Gabriel started a competition for the most obfuscated code? There is a thread in the Chit Chat forum about the most obfuscated "Hello World" application. You might wanna try it!

Oops supposed to be code not quote

I will not try to understand what you did here! Did you and Gabriel started a competition for the most obfuscated code? There is a thread in the Chit Chat forum about the most obfuscated "Hello World" application. You might wanna try it!

It's quite simple really. Differences in consecutive squares are easily calculated. It turns out that in the series, the difference increments by 4 each time. I split the cases between even and odd n:

Odd N:

Sum = 1^2 + (3^2 - 2^2) + (5^2 - 4^2) +...

Sum = 1 + 5 + 9 + 13 + ... (it now becomes obvious this is an arithmetic series)

(use equation for arithmetic series, re-order a few terms and we have:)

Sum = n*(1+2*(n-1)) ( where n = (N+1)/2 )

Even N:

Sum = -[(2^2 - 1^2) + (4^2-3^2) + (6^2 - 5^2) + ...]

Sum = -[3 + 7 + 11 + 15 + ...]

(again, this is an arithmetic series which can be re-written as:)

Sum = -n*(3+2*(n-1)) ( where n = (int)((N+1)/2) )

The equations for both even and odd are very similar, just a negative sign and one constant. That's where I got my equation:

k1*n*(k2+2*(n-1))

And once you understand the math (which I didn't explain b/c I weas looking for a reaction like yours ;) ), the code is perfectly legible! :D

:blush: ooops. I misread it. In that case, the series is not that easy to simplify. LOL!

This is it... updated

/************************************************************************************
* *
* Filename: A2Part5.cpp *
* *
* Login: po028 *
* *
* Assignment No: 2 *
* *
* File Description: To compute and display the series 1 - 2^2 + 3^3 - 4^4 + 5^5 - ..*
* and the result. For example: Enter n = 2 *
* *
* Date Last Modified: 10/04/2005 *
************************************************************************************/

#include <iostream>
using namespace std;

int main()
{
int sign = 1; //for changing signs
int sum = 0; //make sure that the sum is 0
int n; //input
double i; //index

cout << "Enter a value: "; //get input
cin >> n;

if((n <= 6) && (n >= 0)) //check if input is in range
{ //if so do for loop
for(i=1; i<=n; i++) //for loop for calculating series
{
sum += sign * (long int)pow(i, i); //calculate number to power of number
sign *= -1; // swapping the signs
}

cout << "0"; //begin output in the series format
for (int i=1; i<=n; i++) //for loop for output as required
{
if (i%2==0) //if number is even use subtraction sign
{
cout << "-";
}
else //number is then odd, use addition sign
cout << "+";

cout << i << "^" << i; //output number to power of number format
}
cout << "=" << sum << endl; //output the result

}
else //number is not in range
{
cout << "Invalid\n";
}



return 0;
}

It was a cut and paste job plus that teacher thing i doesn't bother me...With the login you guys gotta figure out which Uni i go to first...

int series_sum(const int &n)
{
int sum=0;
for(int i=1,sign=1,p,j;i<=n;sum+=sign*p,sign*=-1,++i)
for(j=1,p=1;j<=i;p*=i,++j);
return sum;
}

IMHO, KevinHall's solution was correct.
That's because you're not reading the posts carefully... ;)

PS: Perhaps you want to run this...

int calcSeries(int N)
{
int k1, k2, n = (N+1)/2;

if (N<1) exit(-1);

if (N%2)
{
k1 = k2 = 1;
}
else
{
k1 = -1;
k2 = 3;
}

return k1*n*(k2+2*(n-1));
}

int powerSums(int N)
{
int sum = 0;
int sign = 1;
for(int i=1; i<=N; i++)
{
sum += sign*(int)pow(i, i);
sign *= -1;
}

return sum;
}

int main()
{
for (int i=1; i<=10; ++i)
{
printf("i = %d\n",i);
printf("[Kevin]\t%d\n", calcSeries(i));
printf("[Cilu]\t%d\n", powerSums(i));
}
return 0;
}

Oh. It was stupid of me. thanks for the explanation.
That's ok. We all do mistakes. After all, we are only programmers... :D