hdu5935 Car（贪心，精度）

Car

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 0    Accepted Submission(s): 0

Problem Description

Ruins is driving a car to participating in a programming contest. As on a very tight schedule, he will drive the car without any slow down, so the speed of the car is non-decrease real number.

Of course, his speeding caught the attention of the traffic police. Police record N positions of Ruins without time mark, the only thing they know is every position is recorded at an integer time point and Ruins started at 0.

Now they want to know the minimum time that Ruins used to pass the last position.

 

Input

First line contains an integer T, which indicates the number of test cases.

Every test case begins with an integers N, which is the number of the recorded positions.

The second line contains N numbers a1, a2, ⋯, aN, indicating the recorded positions.

Limits
1≤T≤100
1≤N≤105
0<ai≤109
ai<ai+1

 

Output

For every test case, you should output 'Case #x: y', where x indicates the case number and counts from 1 and y is the minimum time.

 

Sample Input

1
3
6 11 21

 

Sample Output

Case #1: 4

注意精度问题，假设最后一段为一秒钟完成。往回退，如果可以以当前的速度行驶，就以当前速度行驶，加上对应的时间，如果不行那就把时间加一，更新速度。注意精度即可

#include<iostream>
#include<cmath>
using namespace std;
long long num[100005];
int cal(double e)
{
    if((e)<1e-8)
    {
        return 0;
    }
    else if(e>0)
    {
        return 1;
    }
}
int main()
{
    int cas;
    scanf("%d",&cas);
    int c=0;
    int n;
    while(cas--)
    {
        scanf("%d",&n);
        num[0]=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&num[i]);
        }
        double v=num[n]-num[n-1];
        int ans=1;
        for(int i=n-1;i>=1;i--)
        {
            int dis=num[i]-num[i-1];
            double temp=dis*1.0/(v);
            int flag=cal(temp-(int)temp);
            if(flag==0)
            {
               ans+=(int)temp;    
            }
            else if(flag==1)
            {
                ans+=(int)temp+1;
                v=dis*1.0/(int)(temp+1);
            }
        }
        printf("Case #%d: %d\n",++c,ans);
    }
}