编写程序以生成序列中的前n个项-34,18,10,6,4，

Question

Input is n which is a single integer.
Output consists of the series separated by blank space.
Sample Input:
6
Sample Output:
34 18 10 6 4 3

For this program I had initially declared two variables and initialized them to 34 and 18 respectively. From there on it was easy to calculate the next values and keep printing them. But this was done assuming the series starts from only 34 and the series only consists of 6 terms. This output wasn't accepted as it may have not satisfied all the test cases. So I wrote the following program to find out series based on the value of n entered by the user whereas the program will find out where the series starts and calculate the series from there onward.

#include <stdio.h>
int main()
{
    int a=34,b=18,n,i,sub=16;
    scanf("%d",&n);
    if(n>6)
    {
        for(i=0;i<n-6;i++)
        {
            sub=sub*2;
            a=a+sub;
            b=b+(sub/2);
        }
    }
    printf("%d %d",a,b);
    for(i=0;i<n-2;i++)
    {
        sub=sub/2;
        b=b-sub;
        printf(" %d",b);
    }
    return 0;
}

This way I calculate what the first term of the series is and then start the pattern from there. But it is still not being accepted. Is there some way this code is not satisfying the test cases or is it supposed to be improved?

Answer 1

I suggest using std::generate

Example:

int main()
{
  std::size_t const N = 6;
  std::vector<int> v(N);

  auto start = 0;
  std::cin >> start;

  std::generate(v.begin(), v.end(), [x = start * 2 - 1]() mutable {
    return x = (x / 2 + 1);
  });

  for (auto const& i : v) std::cout << i << ' ';
}

Answer 2

This is quite old, but if anybody stumbles open it one day:

There was several little things that were wrong or could be better in the code :

1 - Don't assume that the user will not enter a number greater than 6 and run the code either way, even if result will be abcde 0 0 0 ... (this is even more important if your code is to be submitted to a machine, as the machine will try a lot more possibilities to break your code)

2 - If you are going to initialize a and b (two first numbers), you'll have to include in your code the possibility of the user only wanting those two numbers.

3 - As you already initialize your first two numbers, don't start your loop at i=0

4 - There could be more things to be pointed out but, mainly, as commented before :

The pattern in the series is that T(n+1) = T(n)/2+1. Your code can be simplified quite a bit.

Here is an easier version of the code (there is always better ones, but this ones is easy to understand at all levels):

Using a while loop :

#include <stdio.h>
  int main() {
    int n, a = 34, i = 2;

    scanf("%d", & n);

    if (n == 1)
      printf("%d", a);
    else {
      printf("%d ", a);
      while (i <= n) {
        a = (a / 2) + 1;
        printf("%d ", a);
        i++;
      }
    }
    return 0;
  }

Another version of the same code using a for loop

#include <stdio.h>
  int main() {
    int n, a = 34, i;

    scanf("%d", & n);

    if (n == 1)
      printf("%d", a);
    else {
      printf("%d ", a);
      for (i = 2; i <= n; i++) {
        a = (a / 2) + 1;
        printf("%d ", a);
      }
    }
    return 0;
  }

I hope this helps !

Answer 3

I think the problem statement is still missing some constraints.

• You either need the starting number like 34 along side number of terms ie 6 form the user.

• Or you should provide more examples to support the fact that starting number is equal to num*num - 2 .

Moreover you can shorten your code length to some extend.

#include <stdio.h>
int main(void) {
    int noOfTerms = 6, number=34;
    for(int i=0;i<noOfTerms;i++){
        printf("%d ",number);
        number = number - (number/2) + 1;
    }
    return 0;
}

Output: 
34 18 10 6 4 3 

Ideone