hdu 5935 Car

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5935

题目描述：

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）

题目分析：

从后往前倒着做，最后一个时间肯定是1s

AC代码：

#include<iostream>#include<cstring>#include<cstdio>#include<cstdlib>#include<cmath>#include<algorithm>using namespace std;const long long int maxn=100005;long long int T,N,cas=1,t,t0,d0,d,time;long long int p[maxn];int main(){    scanf("%lld",&T);    while(T--)    {        scanf("%lld",&N);        for(long long int i=1;i<=N;i++)        {            scanf("%lld",&p[i]);        }        p[0]=0;        t0=1;        t=1;        d0=p[N]-p[N-1];        for(long long int i=N-1;i>=1;i--)        {            d=p[i]-p[i-1];            time=(d*t0+d0-1)/d0;            d0=d;            t0=time;            t+=time;        }        printf("Case #%lld: %lld\n",cas++,t);    }    return 0;}