B

B - Gleb And Pizza

Topic:

   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.

Example:

Input8 477 8 1-7 3 20 2 10 -2 2-3 -3 10 6 25 3 1Output2Input10 840 0 90 0 101 0 11 0 2Output0

题意:

   披萨由半径 r-d 的主要部分组成,原点在中心,在外围宽度 d 的地方有面包皮.香肠的碎片也是圆的.   第i 片的半径为 r[i],圆心为(x[i],y[i])   香肠只有完全在面包皮上才算落在面包皮上.  确认 落在面包皮上的香肠数.

思路:

   根据勾股定理,比较半径,

题解:

#include<bits/stdc++.h>using namespace std;int main(){    int r,d,n,i,x,y,a,s=0;    cin>>r>>d; cin>>n;    for(i=1;i<=n;i++)      {        cin>>x>>y>>a;        if( (x*x)+(y*y)>=(r-d+a)*(r-d+a)  &&  (x*x)+(y*y)<=(r-a)*(r-a)  ) s++;     }    cout<<s;    return 0;}