D. Travel Card

最新推荐文章于 2021-11-12 11:40:30 发布

UMR小豪

最新推荐文章于 2021-11-12 11:40:30 发布

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分类专栏： -----------------动态规划---------------- CF 文章标签： dp codeforces

本文链接：https://blog.youkuaiyun.com/qq_33362864/article/details/54744762

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本文介绍了一种针对公交票务系统的费用优化算法，通过动态规划确定乘客在多次出行中如何选择不同类型的车票以达到最低消费的目标，并实时计算每次出行后的总费用。

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A new innovative ticketing systems for public transport is introduced in Bytesburg. Now there is a single travel card for all transport. To make a trip a passenger scan his card and then he is charged according to the fare.

The fare is constructed in the following manner. There are three types of tickets:

a ticket for one trip costs 20 byteland rubles,
a ticket for 90 minutes costs 50 byteland rubles,
a ticket for one day (1440 minutes) costs 120 byteland rubles.

Note that a ticket for x minutes activated at time t can be used for trips started in time range from t to t + x - 1, inclusive. Assume that all trips take exactly one minute.

To simplify the choice for the passenger, the system automatically chooses the optimal tickets. After each trip starts, the system analyses all the previous trips and the current trip and chooses a set of tickets for these trips with a minimum total cost. Let the minimum total cost of tickets to cover all trips from the first to the current is a, and the total sum charged before is b. Then the system charges the passenger the sum a - b.

You have to write a program that, for given trips made by a passenger, calculates the sum the passenger is charged after each trip.

Input

The first line of input contains integer number n (1 ≤ n ≤ 105) — the number of trips made by passenger.

Each of the following n lines contains the time of trip ti (0 ≤ ti ≤ 109), measured in minutes from the time of starting the system. All tiare different, given in ascending order, i. e. ti + 1 > ti holds for all 1 ≤ i < n.

Output

Output n integers. For each trip, print the sum the passenger is charged after it.

Examples

input

output

20
20
10

input

output

Note

In the first example, the system works as follows: for the first and second trips it is cheaper to pay for two one-trip tickets, so each time 20 rubles is charged, after the third trip the system understands that it would be cheaper to buy a ticket for 90 minutes. This ticket costs 50 rubles, and the passenger had already paid 40 rubles, so it is necessary to charge 10 rubles only.

题意：乘坐电车，一共会有n次乘坐，之后n行会一次给出每次乘坐的时间。问你最少花的钱是多少。每次输出应该支付的钱。

用dp[n]表示第n次旅行一共所要花费的最小钱数。那么对于dp[n]

1.就是花20快钱买咯。

2.和上面的一起能凑个90分钟就凑个90分钟。

3.和上面的一起能凑个一天就凑个1天。

那么当前的值就是这3个值的最小值。

然后看看代码就懂了。

#include <bits/stdc++.h>
using namespace std;
const int MAXN=1e5+7;
int n;
int dp[MAXN];
long long tt[MAXN];
int main()
{
    int i,j;
    scanf("%d",&n);
    for(i=1;i<=n;++i)
    {
        scanf("%I64d",&tt[i]);
    }
   int sum=0;
    for(i=1;i<=n;++i)
    {
        dp[i]=dp[i-1]+20;
        for(j=i-1;j>=1;--j)
        {
            if(tt[i]-tt[j]<90)dp[i]=min(dp[i],dp[j-1]+50);
            else
            if(tt[i]-tt[j]<1440)dp[i]=min(dp[i],dp[j-1]+120);
            else break;
        }
        printf("%d\n",dp[i]-sum);
        sum=dp[i];
    }
    return 0;
}