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.
题目大意：一个分两层的大圆，给出一些小圆，求完全在大圆外层的小圆个数
题解报告：简单模拟，直接\(O(n)\)扫一边即可
#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <cmath>
#define RG register
#define il inline
#define iter iterator
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
using namespace std;
const int N=1e5+5;
int x[N],y[N],r[N];int r1,d,n;
int js(int i){
return x[i]*x[i]+y[i]*y[i];
}
bool check(int i){
int dis=js(i);
int t1=(r1-r[i])*(r1-r[i]),t2=(r[i]+r1-d)*(r[i]+r1-d);
if(dis>=t2 && dis<=t1)return true;
return false;
}
void work()
{
scanf("%d%d%d",&r1,&d,&n);
for(int i=1;i<=n;i++){
scanf("%d%d%d",&x[i],&y[i],&r[i]);
}
int ans=0;
for(int i=1;i<=n;i++){
if(check(i))ans++;
}
printf("%d\n",ans);
}
int main()
{
work();
return 0;
}