Codeforces Round #430 (Div. 2) B. Gleb And Pizza(數論)

B. Gleb And Pizza

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Gleb ordered pizza home. When the courier delivered the pizza, he was very upset, because several pieces of sausage lay on the crust, and he does not really like the crust.

The pizza is a circle of radius r and center at the origin. Pizza consists of the main part — circle of radius r - d with center at the origin, and crust around the main part of the width d. Pieces of sausage are also circles. The radius of the i -th piece of the sausage is ri, and the center is given as a pair (xi, yi).

Gleb asks you to help determine the number of pieces of sausage caught on the crust. A piece of sausage got on the crust, if it completely lies on the crust.

Input

First string contains two integer numbers r and d (0 ≤ d < r ≤ 500) — the radius of pizza and the width of crust.

Next line contains one integer number n — the number of pieces of sausage (1 ≤ n ≤ 105).

Each of next n lines contains three integer numbers xi, yi and ri ( - 500 ≤ xi, yi ≤ 500, 0 ≤ ri ≤ 500), where xi and yi are coordinates of the center of i-th peace of sausage, ri — radius of i-th peace of sausage.

Output

Output the number of pieces of sausage that lay on the crust.

Examples

input

8 4
7
7 8 1
-7 3 2
0 2 1
0 -2 2
-3 -3 1
0 6 2
5 3 1

output

2

input

10 8
4
0 0 9
0 0 10
1 0 1
1 0 2

output

0

Note

Below is a picture explaining the first example. Circles of green color denote pieces of sausage lying on the crust.

題目大意:以原點爲圓心有一個大圓，半徑爲r，現在給你幾個圓和半徑，現在問你有幾個小圓完整的存在與r-d這個區間內，具體見上圖

解題思路:這個題對於每個小圓來說，去check與大圓圓心的距離，然後卡一下d和r-d的長度即可，注意精度

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <cstdlib>
#include <ctime>
using namespace std;
typedef long long LL;

int r,d,n;
bool check(int x,int y,int rr)
{
	double cnt = sqrt(x*x+y*y);
	if(cnt-double(rr)>=double(r-d)&&cnt+double(rr)<=r)
		return true;
	else
		return false;
}
int main()
{
	int sum,i,x,y,rr;
	cin>>r>>d;
	cin>>n;
	sum=0;
	for(i=1;i<=n;i++)
	{
		cin>>x>>y>>rr;
		if(check(x,y,rr))
		{
			sum++;
		}
	}
	cout<<sum<<endl;
}