You can use for loop just for some test and see that every number n has n+1 of 0 and n of 1 and n-1 of 2 and so on.

Solution

int main(){
 ll x;       cin >> x;
 cout << (x+1) * (x+2) / 2 << "\n";

  
  return 0 ;
}

int main()
{
    long long n;
        cin >> n;
        long long ans = ((n - 1) * (n - 2) / 2) + n * 3;
        cout << ans << endl;
    
    return 0;
}

It is 2 pointers problem and it is a popular technique called sliding window.

Solution

int main(){
   
 ll n, d;       cin >> n >> d;
        vector < ll > nums(n);
        for(auto & i : nums)    cin >> i;
        ll sum = 0;
        ll ans = 0;
        ll l = 0, r = 0;
        ll ll1 = 0, rr = 0;
        while(r < n)
        {
            r = max(r, l);
            while(r < n && sum + nums[r] <= d)
            {
                sum += nums[r];
                r++;
            }
            if(ans < r-l)
            {
                ans = r - l;
                ll1 = l;
                rr = r;
            }
            if(sum >= nums[l])
                sum -= nums[l];
            l++;
        }
        cout << ans << "\n";
        for(ll i = ll1; i < rr; i++)
        {
            cout << nums[i] << " ";
        }
        if(ans != 0)
        cout << "\n";    return 0;
 
}

Solution

#include <iostream>
ll n;
bool isgood(ll m){
    return m * (m+1)/2 <= n;
}
 
ll bs(){
    ll l = 0, r = n, best = -1;
    while(l <= r){
        ll m = l + (r-l)/2;
        if(isgood(m)) best = m, l = m + 1;
        else r = m-1;
    } 
    return best;
}
 
 
int main()
{
    lil_codi_vert();
    int t; cin >> t;
    while(t--){
        cin >> n;
        cout << bs() << "\n";
    }
 
    return 0;
}

Solution

    bool isGood(ll x, vector<ll> &nums, ll n, ll t) {
	ll numOfProd = 0;
	for (auto& I : nums) numOfProd += x / I;
	return numOfProd >= t;
}
 
ll BS(vector<ll> &nums, ll n, ll t) {
	ll l = 1, r = 1;
	while (!isGood(r, nums, n, t)) r <<= 1;
	ll best = r;
	while (l <= r) {
		ll mid = l + (r - l) / 2;
		(isGood(mid, nums, n, t) ? r = mid - 1, best = mid : l = mid + 1);
	}
	return best;
}
 
int main() {
	ll n, t;
	cin >> n >> t;
	vector<ll> nums(n);
	for (auto& I : nums) cin >> I;
	cout << BS(nums, n, t) << endl;
	return 0;
}

Solution

  int main(){
 
ll n, q, x;
	cin >> n;
	ll arr[n]{}, prefix[n + 1]{};
	for (int i = 0; i < n; i++) {
		cin >> arr[i];
		prefix[i + 1] = prefix[i] + arr[i];
	}
	cin >> q;
	while (q--) {
		cin >> x;
		cout << lower_bound(prefix, prefix + n + 1, x) - prefix << endl;
	}
	return 0;
} 

Solution

int main() {
        int n;
	cin >> n;
	int arr[n];
	for (auto& I : arr) cin >> I;
	deque<int> dq;
	sort(arr, arr + n, [](int a, int b) {
		return a > b;
	});
	for (int i = 0; i < n; i++) {
		(i % 2 ? dq.push_back(arr[i]) : dq.push_front(arr[i]));
	}
	for (auto& I : dq) {
		cout << I << " ";
	}
	cout << endl;
	return 0;    
}
 

Solution

int main() {
        ll t; cin >> t;
    while(t--){
        ll n, k; cin >> n >> k;
        string s, t; cin >> s >> t;
        map<char,int> m1,m2;
        for(auto&i: s) m1[i]++;
        for(auto&i: t) m2[i]++;
        for(char i = 'a'; i <= 'z'; i++){
            while(m1[i] > m2[i] && m1[i] >= k){
                m1[i+1] += k, m1[i] -= k; 
            }
        }
        if(m1 == m2) cout << "Yes\n";
        else cout << "No\n";
    }
 
    return 0;  
}