Car HDU

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

 

废墟正在开车参加编程比赛。 由于时间非常紧张，他会在没有任何减速的情况下驾驶赛车，因此赛车的速度是非减少的实数。

当然，他的超速驾驶引起了交警的注意。 警方在没有时间标记的情况下记录了N个遗址的废墟位置，他们知道的每一个位置唯一的记录是在整数时间点记录的，废墟从0开始记录。

现在他们想知道遗迹用于通过最后位置的最短时间。

这题是个想法题，速度递增 先算出速度的最大值 速度可以为小数，

当速度不等时，可以略微的下调速度  b=temp/(t+1);

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<queue>
#include<set>
#include<cctype>
using namespace std;
int a[100010];
int main() {
    int t,cas=1,n;
    scanf("%d",&t);
    while(t--) {
        scanf("%d",&n);
        for (int i=0 ; i<n ; i++)
            scanf("%d",&a[i]);
        long long ans=0;
        double b=a[n-1]-a[n-2];
        for (int i=n-1 ; i>=1 ; i--) {
            double  temp=(a[i]-a[i-1])*1.0;
            int t=temp/b;
            ans+=t;
            if (temp/t!=b) {
                ans++;
                b=temp/(t+1);
            }
        }
        printf("Case #%d: %d
",cas++,ans);
    }
    return 0;
}