WEEK 5 B TT's Magic Cat

topic:

Thanks to everyone’s help last week, TT finally got a cute cat. But what TT didn’t expect is that this is a magic cat.
One day, the magic cat decided to investigate TT’s ability by giving a problem to him. That is select nn cities from the world map, and a[i]a[i] represents the asset value owned by the ii-th city.
Then the magic cat will perform several operations. Each turn is to choose the city in the interval [l,r]and increase their asset value by cc. And finally, it is required to give the asset value of each city after qq operations.
Could you help TT find the answer?

input：

The first line contains two integers n,q (1≤n,q≤2⋅105) — the number of cities and operations.
The second line contains elements of the sequence aa: integer numbers a1,a2,…,an(−106≤ai≤106)
Then qq lines follow, each line represents an operation. The ii-th line contains three integers l,r and c (1≤l≤r≤n,−105≤c≤105) for the ii-th operation.

output：

Print nn integers a1,a2,…,an one per line, and ai should be equal to the final asset value of the ii-th city.

example：

Ideas:

Assuming that the length of the array element to be processed is q, if the problem is directly used to process the original array, the time complexity is O (q * n); if the original array is converted into a differential array for processing, the time complexity will be Reduced to O (q + n); and if the original array is converted into a difference array for processing, the original interval modification will be changed to point modification, that is, A [L, R] will become B [X] + = C , B [Y + 1]-= C, the prefix sum of difference array B is the final value of A after modification. In summary, the difference array is used for operation here.

note:

This problem should be handled with long long type, not int class.

Code:

#include <iostream>

using namespace std;

const int N = 1e7;
long long n, k, x, y, c;
long long a[N], b[N], temp[N];

void change()
{
	b[1] = a[1];
	for (long long i = 2; i <= n; i++)
	{
		b[i] = a[i] - a[i - 1];
	}
}

int main()
{
	while (scanf_s("%lld %lld",&n,&k)!=EOF)
	{
		for (long long i = 1; i <= n; i++)
			scanf_s("%lld", &a[i]);
		change();
		for (long long i = 0; i < k; i++)
		{
			scanf_s("%lld %lld %lld", &x, &y, &c);
			b[x] += c;
			b[y + 1] -= c;
		}
		temp[1] = b[1];
		cout << temp[1];
		for (long long i = 2; i <= n; i++)
		{
			temp[i] = temp[i - 1] + b[i];
			cout << ' ' << temp[i];
		}
		cout << endl;
	}
    return 0; 
}