#include<stdio.h>

#include<math.h>

int main()

{

int t,n,k,i;

int sum;

scanf("%d",&n);

int a[n];

for(i=0;i<n;i++)

scanf("%d",&a[i]);

sum=0;

k=n-1;

for(i=0;i<n;i++)

{

sum=sum+(a[k]*pow(10,i));

k–;

}

printf("%d\n",sum);

return 0;

}

Hello destiny_me ,

You can achieve using this way …

int num = 0 ; for(int i=0;i<n;i++){ num = num * 10 + arr[i] ; } cout << num << endl ;

Here is the recursive way for the same:

```
int find_num(int a[], int start, int n, int num)
{
if(start==n)
return num;
num= num*10 + a[start];
return find_num(a,start+1,n,num);
}
```

with the call : `cout<<find_num(a,0,n,0);`

// a is the input array and n is total number in the array and start refers to the starting index i.e. 0 and num, is the number formed at some particular step.

you can just use StringBuilder or StringBuffer and append every number of array and last you can convert string into number by parsing it…