Car HDU - 5935

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 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 .

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

 

 

Input

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

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

The second line contains numbers , , , , indicating the recorded positions.

Limits

 

 

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);

 

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<cmath>
 4 #include<algorithm>
 5 #include<queue>
 6 #include<set>
 7 #include<cctype>
 8 using namespace std;
 9 int a[100010];
10 int main() {
11     int t,cas=1,n;
12     scanf("%d",&t);
13     while(t--) {
14         scanf("%d",&n);
15         for (int i=0 ; i<n ; i++)
16             scanf("%d",&a[i]);
17         long long ans=0;
18         double b=a[n-1]-a[n-2];
19         for (int i=n-1 ; i>=1 ; i--) {
20             double  temp=(a[i]-a[i-1])*1.0;
21             int t=temp/b;
22             ans+=t;
23             if (temp/t!=b) {
24                 ans++;
25                 b=temp/(t+1);
26             }
27         }
28         printf("Case #%d: %d\n",cas++,ans);
29     }
30     return 0;
31 }