You are given an array of integers of size N . You need to print the sum of the elements in the array, keeping in mind that some of those integers may be quite large.
Input
The first line of the input consists of an integerN . The next line contains N space-separated integers contained in the array.
Constraints
1≤N≤10
0≤A[i]≤1010
Sample Input
5
1000000001 1000000002 1000000003 1000000004 1000000005
Output
Print a single value equal to the sum of the elements in the array. In the above sample, you would print5000000015 .
Note: The range of the 32-bit integer is(−231) to (231−1) or [−2147483648,2147483647] .
When we add several integer values, the resulting sum might exceed the above range. You might need to use long long int in C/C++ or long data type in Java to store such sums.
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Input
The first line of the input consists of an integer
Constraints
Sample Input
5
1000000001 1000000002 1000000003 1000000004 1000000005
Output
Print a single value equal to the sum of the elements in the array. In the above sample, you would print
Note: The range of the 32-bit integer is
When we add several integer values, the resulting sum might exceed the above range. You might need to use long long int in C/C++ or long data type in Java to store such sums.
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A Very Big Sum - - Hacker Rank Solution
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int main()
{
long long int a[10],n,i,sum=0;
scanf("%lld",&n);
for(i=0;i<n;i++)
scanf("%lld",&a[i]);
for(i=0;i<n;i++)
sum=sum+a[i];
printf("%lld",sum);
return 0;
}
#include <string.h>
#include <math.h>
#include <stdlib.h>
int main()
{
long long int a[10],n,i,sum=0;
scanf("%lld",&n);
for(i=0;i<n;i++)
scanf("%lld",&a[i]);
for(i=0;i<n;i++)
sum=sum+a[i];
printf("%lld",sum);
return 0;
}