Codeforces Round #178 (Div. 2) / 294A Shaass and Oskols(模拟)

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本文深入探讨了AI音视频处理领域中的关键技术——视频分割与语义识别，通过详细解释这些技术的工作原理、应用场景及实际应用案例，为读者提供了一次全面的技术学习之旅。

Shaass has decided to hunt some birds. There are n horizontal electricity wires aligned parallel to each other. Wires are numbered 1 to nfrom top to bottom. On each wire there are some oskols sitting next to each other. Oskol is the name of a delicious kind of birds in Shaass's territory. Supposed there are a_i oskols sitting on the i-th wire.

Sometimes Shaass shots one of the birds and the bird dies (suppose that this bird sat at the i-th wire). Consequently all the birds on thei-th wire to the left of the dead bird get scared and jump up on the wire number i - 1, if there exists no upper wire they fly away. Also all the birds to the right of the dead bird jump down on wire number i + 1, if there exists no such wire they fly away.

Shaass has shot m birds. You're given the initial number of birds on each wire, tell him how many birds are sitting on each wire after the shots.

Input

The first line of the input contains an integer n, (1 ≤ n ≤ 100). The next line contains a list of space-separated integers a₁, a₂, ..., a_n,(0 ≤ a_i ≤ 100).

The third line contains an integer m, (0 ≤ m ≤ 100). Each of the next m lines contains two integers x_i and y_i. The integers mean that for the i-th time Shaass shoot the y_i-th (from left) bird on the x_i-th wire, (1 ≤ x_i ≤ n, 1 ≤ y_i). It's guaranteed there will be at least y_i birds on the x_i-th wire at that moment.

Output

On the i-th line of the output print the number of birds on the i-th wire.

   Sample test(s)
 

     input
   
5
10 10 10 10 10
5
2 5
3 13
2 12
1 13
4 6

     output
   
0
12
5
0
16

     input
   
3
2 4 1
1
2 2

     output
   
3
0
3

今天又做了一道水题。罪过，罪过。

/*30ms,0KB*/

#include<cstdio>

int a[101];

int main()
{
	int n, m;
	int x, y;
	scanf("%d", &n);
	for (int i = 1; i <= n; ++i)
		scanf("%d", &a[i]);
	scanf("%d", &m);
	while (m--)
	{
		scanf("%d%d", &x, &y);
		if (x == 1)
			a[2] += a[1] - y;
		else if (x == n)
			a[n - 1] += y - 1;
		else
		{
			a[x - 1] += y - 1;
			a[x + 1] += a[x] - y;
		}
		a[x] = 0;
	}
	for (int i = 1; i <= n; ++i)
		printf("%d\n", a[i]);
	return 0;
}