codeforces 702D D. Road to Post Office(数学)

题目链接：

D. Road to Post Office

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Vasiliy has a car and he wants to get from home to the post office. The distance which he needs to pass equals to d kilometers.

Vasiliy's car is not new — it breaks after driven every k kilometers and Vasiliy needs t seconds to repair it. After repairing his car Vasiliy can drive again (but after k kilometers it will break again, and so on). In the beginning of the trip the car is just from repair station.

To drive one kilometer on car Vasiliy spends a seconds, to walk one kilometer on foot he needs b seconds (a < b).

Your task is to find minimal time after which Vasiliy will be able to reach the post office. Consider that in every moment of time Vasiliy can left his car and start to go on foot.

Input

The first line contains 5 positive integers d, k, a, b, t (1 ≤ d ≤ 1012; 1 ≤ k, a, b, t ≤ 106; a < b), where:

• d — the distance from home to the post office;
• k — the distance, which car is able to drive before breaking;
• a — the time, which Vasiliy spends to drive 1 kilometer on his car;
• b — the time, which Vasiliy spends to walk 1 kilometer on foot;
• t — the time, which Vasiliy spends to repair his car.

Output

Print the minimal time after which Vasiliy will be able to reach the post office.

Examples

input

5 2 1 4 10

output

14

input

5 2 1 4 5

output

13

题意：

能开车,能步行，不过开车开一段距离后得修车,问到达目的地的最短时间;

思路：

AC代码：

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┆　　┃　 AC代马 　　┣┓┆
┆　　┃　        　　┏┛┆
┆　　┗┓┓┏━┳┓┏┛ ┆
┆　　　┃┫┫　┃┫┫ ┆
┆　　　┗┻┛　┗┻┛ ┆
************************************************ */  

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <bits/stdc++.h>
#include <stack>

using namespace std;

#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));

typedef  long long LL;

template<class T> void read(T&num) {
    char CH; bool F=false;
    for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
    for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
    F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
    if(!p) { puts("0"); return; }
    while(p) stk[++ tp] = p%10, p/=10;
    while(tp) putchar(stk[tp--] + '0');
    putchar('\n');
}

const LL mod=1e9+7;
const double PI=acos(-1.0);
const int inf=1e9;
const int N=1e5+10;
const int maxn=(1<<8);
const double eps=1e-8;

int main()
{
        LL d,k,a,b,t,ans=0;
        read(d);read(k);read(a);read(b);read(t);

        if(k*a+t<=k*b)
        {
            if(d%k==0)
            {
                ans=d/k*(k*a+t)-t;
            }
            else
            {
                if(d>=k){ans=d/k*(k*a+t)-t;ans=ans+min(d%k*b,t+d%k*a);}
                else ans=d*a;

            }
        }
        else
        {
            ans=min(d,k)*a;
            d=d-min(d,k);
            ans=ans+d*b;
        }
        cout<<ans<<endl;

        return 0;
}