Problem :
You are given an array of n
integers, ar = [ar[0], ar[1], ..., ar[n-1]]
, and a positive integer, k
. Find and print the number of (i, j)
pairs where i < j
and ar[i]
+ ar[j]
is divisible by k
.
For example, ar = [1, 2, 3, 4, 5, 6]
and k = 5
. Our three pairs meeting the criteria are [1, 4] [2, 3]
and [4, 6]
.
Function Description :
Complete the divisibleSumPairs function in the editor below. It should return the integer count of pairs meeting the criteria.
divisibleSumPairs has the following parameter(s):
- n: the integer length of array
ar
- ar: an array of integers
- k: the integer to divide the pair sum by
Input Format :
The first line contains 2
space-separated integers, n
and k
.
The second line contains n
space-separated integers describing the values of ar = [ar[0], ar[1], …, ar[n-1]].
Constraints :
2 <= n <= 100
1 <= k <= 100
1 <= ar[i] <= 100
Output Format :
Print the number of (i, j)
pairs where i < j
and a[i] + a[j]
is evenly divisible by k
.
Sample Input :
6 3
1 3 2 6 1 2
Sample Output :
5
Solution :
#include <bits/stdc++.h>
using namespace std;
vector<string> split_string(string);
// Complete the divisibleSumPairs function below.
int divisibleSumPairs(int n, int k, vector<int> ar) {
int sum = 0, count = 0;
for(int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if (i < j) {
sum = ar[i] + ar[j];
if (sum % k == 0) {
count++;
}
}
sum = 0;
}
}
return count;
}
int main()
{
ofstream fout(getenv("OUTPUT_PATH"));
string nk_temp;
getline(cin, nk_temp);
vector<string> nk = split_string(nk_temp);
int n = stoi(nk[0]);
int k = stoi(nk[1]);
string ar_temp_temp;
getline(cin, ar_temp_temp);
vector<string> ar_temp = split_string(ar_temp_temp);
vector<int> ar(n);
for (int i = 0; i < n; i++) {
int ar_item = stoi(ar_temp[i]);
ar[i] = ar_item;
}
int result = divisibleSumPairs(n, k, ar);
fout << result << "\n";
fout.close();
return 0;
}
vector<string> split_string(string input_string) {
string::iterator new_end = unique(input_string.begin(), input_string.end(), [] (const char &x, const char &y) {
return x == y and x == ' ';
});
input_string.erase(new_end, input_string.end());
while (input_string[input_string.length() - 1] == ' ') {
input_string.pop_back();
}
vector<string> splits;
char delimiter = ' ';
size_t i = 0;
size_t pos = input_string.find(delimiter);
while (pos != string::npos) {
splits.push_back(input_string.substr(i, pos - i));
i = pos + 1;
pos = input_string.find(delimiter, i);
}
splits.push_back(input_string.substr(i, min(pos, input_string.length()) - i + 1));
return splits;
}
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