Codeforces 295A. Greg and Array

Greg has an array a = a1, a2, ..., an and m operations. Each operation looks as: li, ri, di, (1 ≤ li ≤ ri ≤ n). To apply operation i to the array means to increase all array elements with numbers li, li + 1, ..., ri by value di.

Greg wrote down k queries on a piece of paper. Each query has the following form: xi, yi, (1 ≤ xi ≤ yi ≤ m). That means that one should apply operations with numbers xi, xi + 1, ..., yi to the array.

Now Greg is wondering, what the array a will be after all the queries are executed. Help Greg.

Input

The first line contains integers n, m, k (1 ≤ n, m, k ≤ 105). The second line contains n integers: a1, a2, ..., an (0 ≤ ai ≤ 105) — the initial array.

Next m lines contain operations, the operation number i is written as three integers: li, ri, di, (1 ≤ li ≤ ri ≤ n), (0 ≤ di ≤ 105).

Next k lines contain the queries, the query number i is written as two integers: xi, yi, (1 ≤ xi ≤ yi ≤ m).

The numbers in the lines are separated by single spaces.

Output

On a single line print n integers a1, a2, ..., an — the array after executing all the queries. Separate the printed numbers by spaces.

Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams of the %I64d specifier.

numopt[i]表示[i,n]进行了一次操作，所以i的操作次数就是∑numopt[x] (x <=i)

同理对后面的数组累加也是这种操作 

#include <bits/stdc++.h>

typedef long long LL;
using namespace std;
LL n,m,k;
#define SIZE 100005

struct optnode{
  LL l,r,d;
} opt[SIZE];
LL numopt[SIZE] = {0};

LL a[SIZE],c[SIZE] = {0};

int main(){
  // freopen("test.in","r",stdin);
  ios::sync_with_stdio(false);
  cin >> n >> m >> k;
  for (LL  i=1;i<=n;i++){
    cin >> a[i];
  }
  for (LL  i=1;i<=m;i++){
    cin >> opt[i].l >> opt[i].r >> opt[i].d;
  }
  for (LL  i=1;i<=k;i++){
    LL x,y;
    cin >> x >> y;
    numopt[x] ++; numopt[y+1] --;
  }
  LL sum = 0;
  for (LL i=1;i<=m;i++){
    sum += numopt[i];
    opt[i].d *= sum;
  }

  for (LL i=1;i<=m;i++){
    c[opt[i].l] += opt[i].d;
    c[opt[i].r+1] -= opt[i].d;
  }

  sum = 0;
  for (LL i=1;i<=n;i++){
    sum += c[i];
    cout << a[i] + sum << " ";
  }
  return 0;
}