hdu 5935 Car

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

题意：给出n个点的坐标（肯定是从小到大），到达每一个点的时间为整数点，速度只能增不能降。

思路：从后推，让最后一个点的速度最大，前面的在满足不降速的情况下取最大速度。

#include <iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
int  a[100005];
int main()
{
    int n,T,sum;
    cin>>T;
    for(int t=1;t<=T;t++)
    {
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
        scanf("%d",&a[i]);
        double v=(a[n]-a[n-1]);
        sum=0;
        for(int i=n-1;i>=1;i--)
        {
            for(int j=(a[i]-a[i-1])/v;;j++)//枚举速度 j为从a[i]~a[i-1] 的时间。
            {
                double y=(a[i]-a[i-1])*1.0/j;
                if(y<=v)
                {
                    sum+=j;
                    v=y;
                    break;
                }
            }
        }
        cout<<"Case #"<<t<<": "<<sum+1<<endl;
    }
    return 0;
}