CodeForces 908C New Year and Curling

合集 - Codeforces(21)

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题目链接：CodeForces 908C【New Year and Curling】

思路

模拟，考虑到两个圆盘可能出现y值相同且相接的情况，所以在判断当前圆盘的y值时循环的范围从在前圆盘的x值左右浮动2r，依次遍历这个范围内的数组y（存储的是当前已经移动了圆盘中的横坐标为i的圆盘的最大的y值），然后可以通过三角形的边长计算出dis，依次枚举每个x取得到的y中的最大值。

代码

 #include <iomanip>
#include <iostream>
#include <math.h>
using namespace std;
 
#define ll long long
const int N = 2e3 + 10;
 
double l, r;
int x[N];
double y[N], mark[N];
int n;
 
void solve() {
  cin >> n >> r;
  for (int i = 1; i <= n; i++) {
    cin >> x[i];
    // 存储当前情况最多允许x[i]位置的元素下落到哪个位置
    double relay = 0;
    for (int j = max(-r * 2, (double)-x[i] + 1); j <= (int)r * 2; j++) {
      double dis = 4 * r * r - j * j;
      if (j == 0 && y[x[i]] == 0)
        dis = r;
      else if (j == 0)
        dis = 2 * r + y[x[i]];
      else if (y[j + x[i]] == 0)
        dis = r;
      else
        dis = sqrt(dis) + y[j + x[i]];
      
      relay = max(relay, dis);
    }
    y[x[i]] = relay;
    mark[i] = relay;
  }
 
  cout << setprecision(12);
  for (int i = 1; i <= n; i++) {
    cout << mark[i] << " ";
  }
}
 
int main() {
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

posted @ 2024-08-01 09:53 薛定谔的AC 阅读(11) 评论(0) 编辑收藏举报

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CodeForces 908C New Year and Curling

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	#include <iomanip>
	#include <iostream>
	#include <math.h>
	using namespace std;

	#define ll long long
	const int N = 2e3 + 10;

	double l, r;
	int x[N];
	double y[N], mark[N];
	int n;

	void solve() {
	cin >> n >> r;
	for (int i = 1; i <= n; i++) {
	cin >> x[i];
	// 存储当前情况最多允许x[i]位置的元素下落到哪个位置
	double relay = 0;
	for (int j = max(-r * 2, (double)-x[i] + 1); j <= (int)r * 2; j++) {
	double dis = 4 * r * r - j * j;
	if (j == 0 && y[x[i]] == 0)
	dis = r;
	else if (j == 0)
	dis = 2 * r + y[x[i]];
	else if (y[j + x[i]] == 0)
	dis = r;
	else
	dis = sqrt(dis) + y[j + x[i]];

	relay = max(relay, dis);
	}
	y[x[i]] = relay;
	mark[i] = relay;
	}

	cout << setprecision(12);
	for (int i = 1; i <= n; i++) {
	cout << mark[i] << " ";
	}
	}

	int main() {
	int t = 1;
	while (t--) {
	solve();
	}
	return 0;
	}