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 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[];
 int main() {
     int t,cas=,n;
     scanf("%d",&t);
     while(t--) {
         scanf("%d",&n);
         for (int i= ; i<n ; i++)
             scanf("%d",&a[i]);
         long long ans=;
         double b=a[n-]-a[n-];
         for (int i=n- ; i>= ; i--) {
             double  temp=(a[i]-a[i-])*1.0;
             int t=temp/b;
             ans+=t;
             if (temp/t!=b) {
                 ans++;
                 b=temp/(t+);
             }
         }
         printf("Case #%d: %d\n",cas++,ans);
     }
     return ;
 }