HDU 5935

Car

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2701    Accepted Submission(s): 728

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

136 11 21

 

Sample Output

Case #1: 4

 

Source

2016年中国大学生程序设计竞赛（杭州） 

 

题意：

在n个点上有摄像头，进过这几个点的时间必须是整数，你从0秒开始，0点开始。速度只能增加或不变，求用时最少。

POINT：

从最后往前推，最后的速度肯定是V=x[n]-x[n-1] m/s，让最后一段用时1秒，然后用这个速度去比较前面一段，看看能不能整除，不能整除就在(x[n-1]-x[n-2])/V的时间上向上取整。在算出这个速度。重复做这个操作。O(n)

#include<iostream>#include<cstring>#include<cmath>#include<iomanip>using namespace std;#define ll long long int#define inf 0x7fffffffconst int maxn = 1e5 + 10;const double eps = 0.00000001;double dp[maxn];int c[maxn];int main(){int T;int n;int cs = 0;scanf("%d", &T);while (T--){cs++;memset(dp, 0, sizeof(dp));memset(c, 0, sizeof(c));scanf("%d", &n);for (int i = 1; i <= n; i++){scanf("%d", &c[i]);dp[i] = c[i] - c[i - 1];}double Speed = dp[n];int cnt = 1;for (int i = n - 1; i >= 1; i--){double Tmp = dp[i] / Speed;if (fabs(Tmp-(int)Tmp)<=eps){cnt = cnt + (int)Tmp;}else{cnt = cnt + (int)Tmp + 1;Speed = dp[i] / ((int)Tmp + 1);}}printf("Case #%d: %d\n", cs, cnt);}return 0;}