浙江财经大学
信息管理与工程学院

HDU 1008 Elevator

本文由 Ocrosoft 于 2016-09-09 14:41:39 发表

Elevator

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 64191    Accepted Submission(s): 35359

Problem Description
The highest building in our city has only one elevator. A request list is made up with N positive numbers. The numbers denote at which floors the elevator will stop, in specified order. It costs 6 seconds to move the elevator up one floor, and 4 seconds to move down one floor. The elevator will stay for 5 seconds at each stop.
For a given request list, you are to compute the total time spent to fulfill the requests on the list. The elevator is on the 0th floor at the beginning and does not have to return to the ground floor when the requests are fulfilled.
 

Input
There are multiple test cases. Each case contains a positive integer N, followed by N positive numbers. All the numbers in the input are less than 100. A test case with N = 0 denotes the end of input. This test case is not to be processed.
 

Output
Print the total time on a single line for each test case. 
 

Sample Input
	
1 2 3 2 3 1 0
 

Sample Output
	
17 41
 

Author
ZHENG, Jianqiang
 

Solution
出题人表示没有坐过电梯。
题意:每行给出一些整数,第一个数表示后面数字的个数,后面的数字表示要在这一层停。电梯从第0层开始,上升花6s,下降花4s,每次停留5s。如果已经在这一层了,下次还是在这一层,那么再停5s。
也有可能某个人犯贱,到了这一层等到电梯关了才发现自己要到这一层出去…

#include <set>
#include <map>
#include <list>
#include <cmath>
#include <stack>
#include <queue>
#include <ctime>
#include <string>
#include <cstdio>
#include <vector>
#include <cctype>
#include <climits>
#include <sstream>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <functional>
#define ms(a) memset(a,0,sizeof(a))
typedef long long LL;
const LL LINF = 1e17;
const int INF = INT_MAX / 2;
const int MAXN = 150 + 10;
const int MOD = 100000;
int gcd(int a, int b)
{
	if (!b)return a;
	return gcd(b, a%b);
}
using namespace std;
//or2..or2..or2..or2..//
int main()
{
	int n, t;
	while (cin >> n&&n)
	{
		int ans=0, now=0;
		for (int i = 0; i < n; i++)
		{
			cin >> t;
			ans += (t > now ? (t - now) * 6 : (now - t) * 4) + 5;
			now=t;
		}
		cout << ans << endl;
	}
	return 0;
}

 

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