Simple Array Sum
Given an array of integers, find the sum of its elements.
Function Description
Complete the simpleArraySum function in the editor below. It must return the sum of the array elements as an integer.
simpleArraySum has the following parameter(s):
- ar: an array of integers
Sample Input:
5
1 2 3 4 5
Sample Output
15
Explanation
1+2+3+4+5=15
CODE
#include <bits/stdc++.h>
using namespace std;
string ltrim(const string &);
string rtrim(const string &);
vector<string> split(const string &);
/*
* Complete the 'simpleArraySum' function below.
*
* The function is expected to return an INTEGER.
* The function accepts INTEGER_ARRAY ar as parameter.
*/
int simpleArraySum(vector<int> ar) {
int sum=0;
for (int i=0; i<ar.size(); i++) {
sum+=ar[i];
}
return sum;
}
int main()
{
ofstream fout(getenv("OUTPUT_PATH"));
string ar_count_temp;
getline(cin, ar_count_temp);
int ar_count = stoi(ltrim(rtrim(ar_count_temp)));
string ar_temp_temp;
getline(cin, ar_temp_temp);
vector<string> ar_temp = split(rtrim(ar_temp_temp));
vector<int> ar(ar_count);
for (int i = 0; i < ar_count; i++) {
int ar_item = stoi(ar_temp[i]);
ar[i] = ar_item;
}
int result = simpleArraySum(ar);
fout << result << "\n";
fout.close();
return 0;
}
string ltrim(const string &str) {
string s(str);
s.erase(
s.begin(),
find_if(s.begin(), s.end(), not1(ptr_fun<int, int>(isspace)))
);
return s;
}
string rtrim(const string &str) {
string s(str);
s.erase(
find_if(s.rbegin(), s.rend(), not1(ptr_fun<int, int>(isspace))).base(),
s.end()
);
return s;
}
vector<string> split(const string &str) {
vector<string> tokens;
string::size_type start = 0;
string::size_type end = 0;
while ((end = str.find(" ", start)) != string::npos) {
tokens.push_back(str.substr(start, end - start));
start = end + 1;
}
tokens.push_back(str.substr(start));
return tokens;
}
THANKS FOR VISITING
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