The sum of 1 (1-2) (1-2 3) (1-2 3-n)... where even integers are -k and odd integers are k-CodePudding

I want to write a program where a user tells me an integer(n) and i calculate The sum of 1 (1-2) (1-2 3) (1-2 3-n)... where even integers are -k and odd integers are k.

Ive made a function which does that But the sum is never correct. For example for n=2 it should be sum=0 but shows sum=-1 for n=3 should be sum= 2 but i shows sum=3. (Ignore the debugging printfs)

#include <stdio.h>

int athroismaAkolouthias(int n); // i sinartisi me tin opoia ypologizete to athroisma akolouthias 1 (1-2) (1-2 3) (1-2 3-4).....

int main(){
    int n;
    printf("give n: ");
    scanf("%d", &n);
    printf("the sum is %d", athroismaAkolouthias(n));
}

int athroismaAkolouthias(int n){
    int sum1=0, sum2=0,sum=0;
    int i, temp, j;
    for (i=1; i<=n; i  ){
        for (j=1; j<=i; j  ){
            temp=j;
        }
        if (i%2==0){sum=sum-temp; printf("test1 %d%d",sum,temp);}
        else{sum=temp; printf("test2 %d%d",sum,temp);}
    }
    return sum;
}

CodePudding user response：

Your issue is with our loop which iterate with j, it should update the inner_sum based on j even/odd condition as follows:

#include <stdio.h>

int akl(int n) {
  int sum = 0;

  for (int i = 1; i <= n; i  ) {
    int inner_sum = 0;
    for (int j = 1; j <= i; j  ) {
      if (j % 2 == 0) {
        inner_sum -= j;
      } else {
        inner_sum  = j;
      }
    }
    sum  = inner_sum;
  }

  return sum;
}

int main() {
  int n;

  scanf("%d", &n);

  printf("%d\n", akl(n));
}

You only need two variables for sum that I named them inner_sum and sum which shows the sum of each term and sum over all terms.

CodePudding user response：

Suspicious Line: else {sum = temp; ...

Shouldn't you be adding or subtracting to sum every time??
Why are you assigning to it here, without an addition or subtraction?

You also have variables sum, sum1, and sum2. You print sum1 and sum2, but never modify them.

Here's my solution:

// The sum of 1 (1-2) (1-2 3) (1-2 3-n)... where even integers are -k and odd integers are  k.
#include <stdio.h>

int ancho(int n)
{
    int sum=0;
    for(int i=1; i<=n;   i)
    {
        for(int j=1; j<=i;   j)
        {
            sum  = (2*(j%2)-1)*j;
        }
    }
    return sum;
}

int main(void)
{
    int n = 5;
    
    printf("Solution is %d\n", ancho(n));
}

// Solution is 3 for n = 5, 
// because: 1   (1-2)   (1-2 3)   (1-2 3-4)   (1-2 3-4 5) = 
// 1-1 2-2 3 = 3

Output

Success #stdin #stdout 0s 5476KB
Solution is 3

Output

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