勵志用更少的代碼做更高效的表達

---

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

---

由于速度不會減小，且花費時間都是整數，則倒著來
最后兩點之間的速度最大，且時間最小，v=a[n]-a[n-1];
這之前的速度<=之后的速度

注意：車速可以是小數， 小時數只能是整數， 車速每段路程一定增加。

---

代碼展示

#include<bits/stdc++.h> #define maxn 100005 int main() {int t; scanf("%d",&t);int n,sum;int a[maxn];for(int cas=1;cas<=t;cas++) {scanf("%d",&n);for(int i=1;i<=n;i++) scanf("%d",&a[i]);double v=(a[n]-a[n-1]);sum=1;for(int i=n-1;i>=1;i--)for(int j=(a[i]-a[i-1])/v; ;j++) {double y=(a[i]-a[i-1])*1.0/j;if(y<=v) {sum+=j; v=y; break;}}printf("Case #%d: %d\n",cas,sum);} return 0; }

---

日拱一卒，功不唐捐

總結

以上是生活随笔為你收集整理的21行代码AC_HDU 5935 Car【贪心, 精度】的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。