Segment Tree Lazy Propogation

// Pratiyush Mishra

#include <bits/stdc++.h>
#define ull unsigned long long int
#define ll long long int
#define LL_MAX 9223372036854775807
#define pb push_back
#define pf push_front
#define mp make_pair
#define popb pop_back
#define vl vector<ll>
#define pl pair<ll,ll>
#define bs(v, x) binary_search(v.begin(), v.end(), x)
#define mem1(a) memset(a, -1, sizeof(a))
#define popf pop_front
#define p0(x) cout << x << " "
#define p1(x) cout << x << '\n'
#define p2(x, y) cout << x << " " << y << '\n'
#define p3(x, y, z) cout << x << " " << y << " " << z << '\n'
#define printv(v)                   \
  for (ll i = 0; i < v.size(); ++i) \
    cout << v[i] << " ";            \
  cout << '\n'
#define pr1(x) cout << fixed << setprecision(15) << x << '\n'
#define ordered_set tree<pair<ll, ll>, null_type, less<pair<ll, ll>>, rb_tree_tag, tree_order_statistics_node_update>
// ordered_set s ; s.order_of_key(val)  no. of elements strictly less than val
// s.find_by_order(i)  itertor to ith element (0 indexed)
#define mod 1000000007
#define mod1 998244353
#define fio                         \
  ios_base::sync_with_stdio(false); \
  cin.tie(NULL)
#define get(n) \
  ll n;        \
  cin >> n
#define getvec(v, n)         \
  vector<ll> v(n);           \
  for (ll i = 0; i < n; i++) \
    cin >> v[i];
#define getstr(s) \
  string s;       \
  cin >> s
#define all(x) x.begin(), x.end()
#define countBits(x) __builtin_popcount(x)
using namespace std;


// Solution to the problem: https://codeforces.com/edu/course/2/lesson/5/1/practice/contest/279634/problem/A

ll n, q;
ll l, r, pos, val;
const ll M = 100000 * 4 + 50;
ll tree[M], lazy[M], v[M];

void build(int tl, int tr, ll node)
{
    if (tl == tr)
    {
        tree[node] = v[tl];
        return;
    }
    int tm = (tl + tr) / 2;
    build(tl, tm, 2 * node);
    build(tm + 1, tr, 2 * node + 1);
    tree[node] = (tree[2 * node] + tree[2 * node + 1]);
}

void apply(int tl, int tr, ll node, ll val)
{
    tree[node] += (val * (tr - tl + 1));
    if (tl != tr)
        lazy[node] += val;
}

void passdown(int tl, int tr, ll node)
{
    int tm = (tl + tr) / 2;
    apply(tl, tm, 2 * node, lazy[node]);
    apply(tm + 1, tr, 2 * node + 1, lazy[node]);
    lazy[node] = 0;
}

void update(int tl, int tr, ll node)
{
    if (tl > r || tr < l)
        return;
    if (tl >= l && tr <= r)
    {
        apply(tl, tr, node, val);
        return;
    }
    passdown(tl, tr, node);
    int tm = (tl + tr) / 2;
    update(tl, tm, 2 * node);
    update(tm + 1, tr, 2 * node + 1);
    tree[node] = (tree[2 * node] + tree[2 * node + 1]);
}

ll query(int tl, int tr, int node)
{
    if (tl > r || tr < l)
        return 0;
    if (tl >= l && tr <= r)
        return tree[node];
    passdown(tl, tr, node);
    int tm = (tl + tr) / 2;
    ll ans = query(tl, tm, 2 * node) + query(tm + 1, tr, 2 * node + 1);
    return ans;
}

void mainSolve()
{
    cin >> n >> q;
    for (int i = 1; i <= n; i++)
        v[i] = 0;
    build(1, n, 1);
    while (q--)
    {
        int t;
        cin >> t;
        if (t == 1)
        {
            cin >> l >> r >> val;
            ++l;
            update(1, n, 1);
        }
        else
        {
            cin >> l;
            ++l;
            r = l;
            cout << query(1, n, 1) << endl;
        }
    }
}

int main()
{
    fio;
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif
    //get(t);
    ll t = 1;
    while (t--)
    {
        mainSolve();
    }
    return 0;
}