hdu 5078 Osu!(鞍山现场赛)
Osu!
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 20 Accepted Submission(s): 15
Special Judge
Now, you want to write an algorithm to estimate how diffecult a game is.
To simplify the things, in a game consisting of N points, point i will occur at time ti at place (xi, yi), and you should click it exactly at ti at (xi, yi). That means you should move your cursor
from point i to point i+1. This movement is called a jump, and the difficulty of a jump is just the distance between point i and point i+1 divided by the time between ti and ti+1. And the difficulty of a game is simply the difficulty
of the most difficult jump in the game.
Now, given a description of a game, please calculate its difficulty.
For each test case, the first line contains an integer N (2 ≤ N ≤ 1000) denoting the number of the points in the game. Then N lines follow, the i-th line consisting of 3 space-separated integers, ti(0 ≤ ti < ti+1 ≤ 106),
xi, and yi (0 ≤ xi, yi ≤ 106) as mentioned above.
Your answer will be considered correct if and only if its absolute or relative error is less than 1e-9.
2
5
2 1 9
3 7 2
5 9 0
6 6 3
7 6 0
10
11 35 67
23 2 29
29 58 22
30 67 69
36 56 93
62 42 11
67 73 29
68 19 21
72 37 84
82 24 98
9.2195444573
54.5893762558HintIn memory of the best osu! player ever Cookiezi.
求最大难度,难度为相邻两点的距离除以时间差。
代码:
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <math.h>
using namespace std;
double a[10000];
double b[10000];
int ti[10000];
double dis(int i,int j)
{
return sqrt((a[i]-a[j])*(a[i]-a[j])+(b[i]-b[j])*(b[i]-b[j]));
}
int main()
{
int t,n;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
double ans=0; for(int i=0;i<n;i++)
{
scanf("%d%lf%lf",&ti[i],&a[i],&b[i]);
}
for(int i=1;i<n;i++)
{
ans=max(ans,(dis(i-1,i)/(ti[i]-ti[i-1])));
}
printf("%.10f\n",ans);
}
return 0;
}
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