Soluție HackerRank pentru Summing the N series, subdomeniul Fundamentals, în C++14. Include cerința formatată, exemple, explicația pașilor și cod sursă.
- Problemă: Summing the N series
- Domeniu: Fundamentals
- Limbaj: C++14
Challenge: Summing the N series
Subdomeniu: Fundamentals (fundamentals)
Scor cont: 20.0 / 20
Submission status: Accepted
Submission score: 1.0
Submission ID: 464723435
Limbaj: cpp14
Link challenge: https://www.hackerrank.com/challenges/summing-the-n-series/problem
Cerință
There is a sequence whose n^{text{th}} term is
T_n = n^2 - (n-1)^2
Evaluate the series
S_n = T_1 + T_2 + T_3 + ·s + T_n
Find S_n bmod (10^9+7).
Example
n = 3
The series is 1^2-0^2 + 2^2-1^2 + 3^2-2^2 = 1 + 3 + 5 = 9.
Function Description
Complete the *summingSeries* function in the editor below.
*summingSeries* has the following parameter(s):
- *int n:* the inclusive limit of the range to sum
Returns
- *int:* the sum of the sequence, modulo (10^9+7)
Input Format
The first line of input contains t, the number of test cases.
Each test case consists of one line containing a single integer n.
Constraints
+ 1 ≤ t ≤ 10
+ 1 ≤ n ≤ 10^16
Cod sursă
#include <bits/stdc++.h>
using namespace std;
string ltrim(const string &);
string rtrim(const string &);
int summingSeries(long n) {
return ((n%1000000007)*(n%1000000007))%1000000007;
}
int main()
{
ofstream fout(getenv("OUTPUT_PATH"));
string t_temp;
getline(cin, t_temp);
int t = stoi(ltrim(rtrim(t_temp)));
for (int t_itr = 0; t_itr < t; t_itr++) {
string n_temp;
getline(cin, n_temp);
long n = stol(ltrim(rtrim(n_temp)));
int result = summingSeries(n);
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;
}