PROBLEM LINK:
Setter: srikkanth_adm
Testers: tabr
Editorialist: aryanag_adm
DIFFICULTY:
592
PREREQUISITES:
None
PROBLEM:
You are given a single integer N. Output the number of ordered pairs of strictly positive integers i, j such that i + j = N
EXPLANATION:
Output N-1.
For any 1 \leq i < N, \ i+j = N implies that j must be N - i. Therefore, we have exactly N-1 ordered pairs.
TIME COMPLEXITY:
O(1)
SOLUTION:
Editorialist's Solution
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
//#include <sys/resource.h>
#define double long double
#define initrand mt19937 mt_rand(time(0));
#define rand mt_rand()
#define MOD 1000000007
#define INF 1000000000
#define mid(l, u) ((l+u)/2)
#define rchild(i) (i*2 + 2)
#define lchild(i) (i*2 + 1)
#define mp(a, b) make_pair(a, b)
#define lz lazup(l, u, i);
#define ordered_set tree<pair<int, int>, null_type,less<pair<int, int>>, rb_tree_tag,tree_order_statistics_node_update>
using namespace std;
using namespace __gnu_pbds;
signed main(){
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
int n;
cin>>n;
cout<<(n-1)<<endl;
}