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

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：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 longconst 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;}