HDOJ 5935 Car【模拟】

Car

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 256    Accepted Submission(s): 92

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）

题意是每段路都必须在整数秒内走完 匀速 然后每段速度是非递减的 速度无上限 不得不说 做题时思维太狭隘太局面了 将整个路程分段了 其实从整体来看 将视点提高从全局看 既然速度非递减 那么最后一段路程的速度肯定是最大的啊 然后从后往前推就好了  至于精度听说double已经满足不了了 需要用分数表示速度。

#include <iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#define maxn 100010
using namespace std;
int a[maxn];
int gcd(int a,int b){
    return a%b==0?b:gcd(b,a%b);
}
int main()
{
    int t,n,va,vb,ans,cnt=0;
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        a[0]=0;
        ans=0;
        for(int i=1;i<=n;++i){
            scanf("%d",&a[i]);
            a[i-1]=a[i]-a[i-1];
        }
        va=a[n-1];
        vb=1;
        ans=1;
        for(int i=n-2;i>=0;--i){
            int tem=ceil(a[i]*vb*1.0/va);//向上取整
            ans+=tem;
            vb=tem;
            tem=gcd(a[i],tem);
            va=a[i]/tem;
            vb=vb/tem;
        }
        printf("Case #%d: %d\n",++cnt,ans);
    }
    return 0;
}