hdu 5078
Osu!
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 58 Accepted Submission(s): 41
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
54.5893762558鞍山现场赛第签到题,超级水!!
求最大难度,难度为相邻两点的距离除以时间差。#include <stdio.h>
#include <string.h>
#include <iostream>
#include <math.h>
#include <algorithm>
#include <vector>
#include <map>
#include <queue>
#include <stack>
#define PI acos(-1.0)
#define eps 1e-8
#define LL long long
#define moo 1000000007
#define INF -999999999
using namespace std;
long long a[1005],b[1005],c[1005];
double d[1005];
int main()
{
int t;
int n;
scanf("%d",&t);
while(t--)
{ scanf("%d",&n);
double ans=0;
for(int i=0;i<n;i++)
{
scanf("%I64d%I64d%I64d",&a[i],&b[i],&c[i]);
}
for(int i=1;i<n;i++)
{
d[i]=sqrt((b[i]-b[i-1])*(b[i]-b[i-1])+(c[i]-c[i-1])*(c[i]-c[i-1]))/(a[i]-a[i-1]);
ans=max(d[i],ans);
}
printf("%.10f\n",ans);
}
}
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