hdu5935 Car 贪心卡精度-优快云博客

本文链接：https://blog.youkuaiyun.com/EventQueue/article/details/52995941

本文介绍了一个编程问题，即如何计算参赛者在不减速的情况下通过最后一个位置所需的最短时间。问题中考虑了警察记录的不同位置，并提供了解决方案，采用了一种贪心策略，确保速度递减从而获得整体最小时间。

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Car

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 393 Accepted Submission(s): 143

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

贪心策略，倒着看，从后往前，后面的速度尽可能的大，这样才能保证总的时间最短

那么就确定了最后一段的时间为1，这样最后一段的速度就最大化了，把这个速度往前面传，用前一段距离除以后一段速度，得到时间，如果得到的时间不是整数，要向上取整（因为要保证前面的速度比后面的速度小），然后再用这个时间算当前段的实际速度，继续把新的速度往前面传……

本来想用double表示速度，ceil向上取整，结果WA，可能是精度损失，换成分数就过了，真坑

#include <cstdio>
#include <cstring>
#include <iostream>
#include <cmath>

using namespace std;

typedef long long ll;

int T, n, kase, ans, t;
int a[100005];

int main()
{
	cin >> T;
	kase = 0;
	while (T--) {
		scanf("%d", &n);
		for (int i = 1; i <= n; i++) {
			scanf("%d", &a[i]);
		}
		a[0] = 0;
		ll fenzi, fenmu;
		ans = 0;
		for (int i = n - 1; i >= 0; i--) {
			if (i == n - 1) {
				t = 1;
				fenzi = a[i + 1] - a[i];
				fenmu = 1;
				ans += t;
			}
			else {
				int dis = a[i + 1] - a[i];
				fenmu *= dis;
				swap(fenmu, fenzi);
				if (fenzi % fenmu) {
					t = fenzi / fenmu + 1;
				}
				else {
					t = fenzi / fenmu;
				}
				ans += t;
				fenzi = a[i + 1] - a[i];
				fenmu = t;
			}
		}
		printf("Case #%d: %d\n", ++kase, ans);
	}
	return 0;
}