CodeForces - 908C (暴力枚举)
C. New Year and Curling
time limit per test: 2 seconds
memory limit per test: 256 megabytes
input: standard input
output: standard output
Carol is currently curling.
She has n disks each with radius r on the 2D plane.
Initially she has all these disks above the liney = 10100.
She then will slide the disks towards the line y = 0 one by one in order from 1 to n.
When she slides the i-th disk, she will place its center at the point (xi, 10100.
). She will then push it so the disk’s y coordinate continuously decreases, and x coordinate stays constant. The disk stops once it touches the line y = 0 or it touches any previous disk. Note that once a disk stops moving, it will not move again, even if hit by another disk.
Compute the y-coordinates of centers of all the disks after all disks have been pushed.
Input
The first line will contain two integers n and r (1 ≤ n, r ≤ 1 000), the number of disks, and the radius of the disks, respectively.
The next line will contain n integers x1, x2, …, xn (1 ≤ xi ≤ 1 000) — the x-coordinates of the disks.
Output
Print a single line with n numbers. The i-th number denotes the y-coordinate of the center of the i-th disk. The output will be accepted if it has absolute or relative error at most 10 - 6.
Namely, let’s assume that your answer for a particular value of a coordinate is a and the answer of the jury is b. The checker program will consider your answer correct if for all coordinates.
Example
input
6 2
5 5 6 8 3 12
output
2 6.0 9.87298334621 13.3370849613 12.5187346573 13.3370849613
Note
The final positions of the disks will look as follows:
In particular, note the position of the last disk.
问题分析
题意:从上面往下落球,如果球达到y坐标为0时或者碰到其他球都会停止。给出球的半径和每个球的横坐标,求最后每个球的圆心Y坐标。
每当下落一个球,就去判断它会和哪个球碰到,枚举可以和它发生碰撞的球的最高y坐标加上高度即可(h = sqrt(4*r*r-x*x)勾股定理)。
好,下面是AC code(^_^)
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <queue>
#include <cmath>
using namespace std;
const int N = 1500;
int v[N];
double ans[N];
int main()
{
// freopen("in.txt","r",stdin);
int n,r;
cin>>n>>r;
for(int i = 0; i < n; i++)
cin>>v[i];
int t = (r+r)*(r+r);
for(int i = 0; i < n; i++)
{
int x;double temp = r;
for(int j = 0; j < i; j++)
{
x = fabs(v[i]-v[j]);
if(x<=2*r)//碰撞条件
{
temp = max(ans[j]+sqrt(fabs(t-x*x)),temp);
}
}
ans[i] = temp;
printf("%.10f ",ans[i]);
}
return 0;
}
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