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.
思路：
水题，搞清楚半径关系就行。
代码：
#include <cstdio>
#include <iostream>
using namespace std;
int main()
{
int r,d;
cin>>r>>d;
int n;
cin>>n;
int sum = 0;
while(n--)
{
int x,y,R;
scanf("%d %d %d",&x,&y,&R);
if((x*x)+(y*y)>=(r-d+R)*(r-d+R) && (x*x)+(y*y)<=(r-R)*(r-R))sum++;
}
cout<<sum;
return 0;
}
