Codeforces 964B

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分类专栏： Codeforces Codeforces下分之路

本文链接：https://blog.youkuaiyun.com/qq_36720186/article/details/80087462

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探讨了一种消息阅读策略，通过分析消息的成本变化与额外收益的关系来最大化收益。根据成本变化的正负，确定最佳阅读时间。

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B. Messages

time limit per test：

1 second

memory limit per test：

256 megabytes

input：

standard input

output：

standard output

There are n incoming messages for Vasya. The i-th message is going to be received after ti minutes. Each message has a cost, which equals to A initially. After being received, the cost of a message decreases by B each minute (it can become negative). Vasya can read any message after receiving it at any moment of time. After reading the message, Vasya's bank account receives the current cost of this message. Initially, Vasya's bank account is at 0.

Also, each minute Vasya's bank account receives C·k, where k is the amount of received but unread messages.

Vasya's messages are very important to him, and because of that he wants to have all messages read after T minutes.

Determine the maximum amount of money Vasya's bank account can hold after T minutes.

Input

The first line contains five integers n, A, B, C and T (1 ≤ n, A, B, C, T ≤ 1000).

The second string contains n integers ti (1 ≤ ti ≤ T).

Output

Output one integer — the answer to the problem.

Examples

input

Copy

4 5 5 3 5
1 5 5 4

output

Copy

input

Copy

5 3 1 1 3
2 2 2 1 1

output

Copy

input

Copy

5 5 3 4 5
1 2 3 4 5

output

Copy

Note

In the first sample the messages must be read immediately after receiving, Vasya receives A points for each message, n·A = 20 in total.

In the second sample the messages can be read at any integer moment.

In the third sample messages must be read at the moment T. This way Vasya has 1, 2, 3, 4 and 0 unread messages at the corresponding minutes, he gets 40 points for them. When reading messages, he receives (5 - 4·3) + (5 - 3·3) + (5 - 2·3) + (5 - 1·3) + 5 = - 5 points. This is 35 in total.

题意

有n个信息，第i个信息会在ti分收到。每个信息有初始价值A，收到后不读价值会每分钟减少B，读了你会收到对应的价值。此外你每分钟都会收到C*k的价值，其中k是你收到但没读的信的个数。每分钟你可以读无限封信，所有信必须在T时刻之前读完，问你最后你能获得的最大价值是多少？

思路

第一眼看到感觉挺复杂，以为是个DP，但是仔细一想其实很简单。

对于每一封信，如果收到了不读的话，每分钟会少B，多C，那就只要看B和C的大小了。

如果B>C，当然是一收到就读，不然会每分钟减少(B-C)

如果B<C，当然是在T时刻再读，这样每分钟都会增加(C-B)

如果B=C，收到之后任何时刻读都可以。

因此结论就是

B<=C时，ans = A * n

B>C时，ans = [ A + (C - B) * (T - ti) ] 的求和

AC代码

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<vector>
#include<set>
#include<string>
#include<sstream>
#include<cctype>
#include<map>
#include<stack>
#include<queue>
#include<list>
#include<cstdlib>
#include<ctime>
using namespace std;
#define INF 0x3f3f3f3f
typedef long long ll;

int s[1010];

int main()
{
//    freopen("input.txt", "r", stdin);
//    freopen("output.txt", "w", stdout);
    int n, a, b, c, t, ans = 0;
    scanf("%d%d%d%d%d", &n, &a, &b, &c, &t);
    for(int i = 0; i < n; i++)
    	scanf("%d", &s[i]);
    if(b >= c)
    	ans = n * a;
    else
    {
    	sort(s, s + n);
    	int x = c - b;
    	for(int i = 0; i < n; i++)
    	{
    		ans = ans + a + x * (t - s[i]);
    	}
    }
    printf("%d\n", ans);
    return 0;    
}