296C - Greg and Array - CodeForces Solution

/*
    Author : Krish Italiya
    Date : 20-04-2023
*/
#include <bits/stdc++.h>
using namespace std;
#define ull unsigned long long 
#define ll long long int
#define vll vector<ll>
#define UB(v,x) distance(v.begin(),upper_bound(all(v),(x)))
#define LB(v,x) distance(v.begin(),lower_bound(all(v),(x)))
#define vpll vector<pair<ll,ll>>
#define vvll vector<vector<ll>>
#define pll pair<ll, ll>
#define mpll map<ll, ll>
#define ld long double
// #define float double
#define f(i, x, n) for (int i = x; i < n; i++)
#define rf(i, x, n) for (int i = x; i >= n; i--)
#define pb push_back
#define all(a) (a).begin(), (a).end()
#define rall(a) (a).rbegin(), (a).rend()
#define sz(a)   (ll)a.size()
#define pi (3.141592653589)
#define min3(a,b,c) min(a,min(b,c))
#define min4(a,b,c,d) min(a,min(b,min(c,d)))
#define max3(a,b,c) max(a,max(b,c))
#define max4(a,b,c,d) max(a,max(b,max(c,d)))
#define max5(a,b,c,d,e) max(a,max(b,max(c,max(d,e))))
#define min5(a,b,c,d,e) min(a,min(b,min(c,min(d,e))))
bool isprime(ll number){if (number <= 1)return false;if (number == 2)return true;if (number % 2 == 0) return false;int boundary = (int)floor(sqrt(number));for (int i = 3; i <= boundary; i += 2)if (number % i == 0)return false;return true;}
ll gcd(ll x,ll y){if(y==0)return x;return gcd(y,x%y);}
ll exponentiation(long long a, long long b, long long c) {
      long long ans = 1;
      for(int i = 1;i <= b;i++) {
          ans *= a;                             //multiplying a, b times.
          ans %= c;
      }
  return ans;
}

#define slow ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL)
// #define mod  1000000007
#define mod 10000000007
#define endl '\n'

void solve(){
  ll n,m,k;
  cin>>n>>m>>k;
  vll v(n+2);
  f(i,1,n+1)cin>>v[i];
  vector<vector<ll>> ops;
  f(i,0,m){
    ll l,r,d;
    cin>>l>>r>>d;
    ops.push_back({l,r,d});
  }
  ll qs[m+2] = {0};
  f(i,0,k){
    ll x,y;
    cin>>x>>y;
    qs[x] += 1;
    qs[y+1] -= 1;
  }
  f(i,1,m+1){
    qs[i] += qs[i-1];
  }
  // f(i,1,m+1){
  //   cout<<qs[i]<<" ";
  // }cout<<endl;
  ll add[n+2] = {0};
  f(i,1,m+1){
    ll l = ops[i-1][0];
    ll r = ops[i-1][1];
    ll d = ops[i-1][2];

    // cout<<l<<" "<<r<<" "<<d<<endl;
    add[l] += qs[i]*d;
    add[r+1] -= qs[i]*d;
  }


  f(i,1,n+2){
    add[i] += add[i-1];
  }

  f(i,1,n+1){
    v[i] += add[i];
  }

  f(i,1,n+1){
    cout<<v[i]<<" ";
  }cout<<endl;

  


}     





ll j=1;
int main()
{
    slow;
    ll tt = 1;
    // cin>>tt;    
    while (tt--)
    {
        // cout<<"Case "<<j<<": "<<endl;
        // cout<<"Scenario #"<<j<<":"<<endl;
        solve();
        j++;
    }
    return 0;
}