i have to convert an array of integers into number. for the given code if a test case is taken as 1 0 7 6 2 for an array of 5 numbers output is 10761

general
1

#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;
}

2

Just don’t use the pow function if your result will not be a floating type. In this case is recommended to write your own function (or in this case to use another variable, let’s say power, which is multiplied by 10 every loop).

3

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 ;
4

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.

5

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