HDU5935-Car

Car

                                                                             Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
                                                                                                       Total Submission(s): 1296    Accepted Submission(s): 395

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
  

 

Source

2016年中国大学生程序设计竞赛（杭州）

 

题意：一辆车，从t=0开始走，速度只能递增，可为小数。警察在t为整数的时候记录了N次车的位置(整数)，问到达最后一个位置时这辆车总共开了多久

解题思路：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，除会掉精度需要想办法避免除法

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>

using namespace std;

#define LL long long
const int INF=0x3f3f3f3f;

int n;
int a[100005],b[100005],c[100005];

int main()
{
    int t,cas=0;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        a[0]=0;
        for(int i=1; i<=n; i++) scanf("%d",&a[i]);
        for(int i=1; i<=n; i++) b[i]=a[i]-a[i-1];
        memset(c, 0, sizeof c);
        c[n] = 1;
        int ans=1;
        for(int i=n-1; i>=1; i--)
        {
            c[i] = c[i + 1] * b[i] / b[i + 1];
            while((LL)b[i] * c[i + 1] > (LL)b[i + 1] * c[i]) c[i]++;
            ans+=c[i];
        }
        printf("Case #%d: %d\n",++cas,ans);
    }
    return 0;
}