B - Counterclockwise Rotation(计算几何)

B - Counterclockwise Rotation(计算几何)

给定直角坐标求逆时针旋转$d$

可以转极坐标。

一开始没注意 counterclockwise 是逆时针的意思qwq

$(x,y)\rightarrow (r\cos\alpha ,r \sin \alpha),r=\sqrt{x^2+y^2},\alpha =\arctan(\dfrac{y}{x})$

cpp的话用$atan2(y,x)$ 函数比$atan(y/x)$

因为$atan2$ 支持第1到4象限，且包含$x=0$的特殊情况。

// Problem: B - Counterclockwise Rotation
// Contest: AtCoder - AtCoder Beginner Contest 259
// URL: https://atcoder.jp/contests/abc259/tasks/abc259_b
// Memory Limit: 1024 MB
// Time Limit: 2000 ms
// Date: 2022-07-13 15:26:41
// --------by Herio--------

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull; 
const int N=1e3+5,M=2e4+5,inf=0x3f3f3f3f,mod=1e9+7;
const int hashmod[4] = {402653189,805306457,1610612741,998244353};
#define mst(a,b) memset(a,b,sizeof a)
#define db double
#define PII pair<int,int>
#define PLL pair<ll,ll>
#define x first
#define y second
#define pb emplace_back
#define SZ(a) (int)a.size()
#define rep(i,a,b) for(int i=a;i<=b;++i)
#define per(i,a,b) for(int i=a;i>=b;--i)
#define IOS ios::sync_with_stdio(false),cin.tie(nullptr)
void Print(int *a,int n){
  for(int i=1;i<n;i++)
    printf("%d ",a[i]);
  printf("%d\n",a[n]); 
}
template <typename T>    //x=max(x,y)  x=min(x,y)
void cmx(T &x,T y){
  if(x<y) x=y;
}
template <typename T>
void cmn(T &x,T y){
  if(x>y) x=y;
}
const  double pi = acos(-1.0);
int main(){
  int x,y,d;cin>>x>>y>>d;
   double r = sqrt(x*x+y*y);
   double p = atan2(y,x);
   //printf("%f---\n",atan2(0,0));
  p+=d*pi/180;
  printf("%.10f %.10f\n",r*cos(p),r*sin(p));
  return 0;
}