# Codeforces 964-B. Messages（思维） 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 nABC 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

```20
```

input Copy

```5 3 1 1 3
2 2 2 1 1
```

output Copy

```15
```

input Copy

```5 5 3 4 5
1 2 3 4 5
```

output Copy

```35
```

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 1234 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.

``````#include<bits/stdc++.h>
#define LL long long
using namespace std;

const int MAX = 1e3+10;
const int INF = 0x3fffffff;

struct node {
int value;
int time;
}p[MAX];

int main(){
int n,a,b,c,t;
while(scanf("%d%d%d%d%d",&n,&a,&b,&c,&t)!=EOF){
memset(p,0,sizeof(p));
for(int i=0;i<n;i++){
p[i].value = a;
scanf("%d",&p[i].time);
p[i].value -= (t-p[i].time)*b;
}

if(b>=c){
cout<<n*a<<endl;
}
else{
int te = 0;
for(int i=0;i<n;i++){
te += p[i].value;
}
int sum = 0;
for(int i=0;i<n;i++){
sum += (t - p[i].time) * c;
}
cout<<sum + te<<endl;
}
}
return 0;
}``````
原文作者：B树
原文地址: https://blog.csdn.net/l18339702017/article/details/79983381
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