codeforces-702D-Road to Post Office

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

Note

In the first example Vasiliy needs to drive the first 2 kilometers on the car (in 2 seconds) and then to walk on foot 3 kilometers (in 12 seconds). So the answer equals to 14 seconds.

In the second example Vasiliy needs to drive the first 2 kilometers on the car (in 2 seconds), then repair his car (in 5 seconds) and drive 2 kilometers more on the car (in 2 seconds). After that he needs to walk on foot 1 kilometer (in 4 seconds). So the answer equals to 13 seconds.

这是一个sabi推公式的题
首先算出来每停一次所要话费的时间，然后会判断一下距离d,和所要停车修车的距离k的大小，如果d小于k，那么还停个屁啊，直接开车走咯，然后去算就好了，其中要判断一下停车修车的时间和走路话费的时间的大小

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;typedef long long ll;int main(){    ll d,k,a,b,t;    while(scanf("%lld%lld%lld%lld%lld",&d,&k,&a,&b,&t)!=EOF)    {        ll tt=a*k+t;        ll sum=0;        if(d<=k)        {             sum=d*a;        }        else        {            if(tt<b*k)            {                sum=(d/k)*a*k+t*((d)/k-1)+min( (d%k)*a+t,(d%k)*b);            }            else            {                sum=a*k+b*(d-k);            }        }        printf("%I64d\n",sum);    }    return 0;}