# Large sum

Work out the first ten digits of the sum of   numbers.
Input Format
First line contains , next  lines contain a 50 digit number each.
Constraints
Output Format
Print only first 10 digit of the final sum
Sample Input
`53710728753390210279879799822083759024651013574025046376937677490009712648124896970078050417018260538743249861995247410594742333095130581237266173096299194221336357416157252243056330181107240615490825023067588207539346171171980310421047513778063246676`
Sample Output
`2728190129`
Explanation
Summing the numbers we get , first 10 digits are.
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PROGRAM
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#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int main() {

/* Enter your code here. Read input from STDIN. Print output to STDOUT */

int n,i,a[10],k=0;
scanf(“%d”,&n);
char **arr = (char **)malloc(n * sizeof(char *));
for (i=0; i<n; i++)
{
arr[i] = (char *)malloc(50 * sizeof(char));
scanf(“%s”,arr[i]);
}
long long int bal=0;
for(i=49;i>=10;i–)
{
for(int j=0;j<n;j++)
{
int t=arr[j][i]-48;
bal=bal+t;
}
bal=bal/10;
}
for(i=9;i>=0;i–)
{
for(int j=0;j<n;j++)
{
int t=arr[j][i]-48;
bal=bal+t;
}
a[k]=bal%10;
k++;
bal=bal/10;
}
printf(“%lld”,bal);
/* count number of digits */
int c=0;
while(bal>0)
{
bal=bal/10;
c++;
}
for(i=9;i>=c;i–)
printf(“%d”,a[i]);

return 0;
}