思路：由于车的速度肯定大于人的，那么第一段肯定是开车最优，然后分三种情况讨论，分别是剩下的都走路，一直开车到最后一段<k的时候走路，或者全程开车
#include<bits/stdc++.h>
using namespace std;
#define LL long long
LL d,k,a,b,t;
int main()
{
LL ans = 0;
scanf("%lld%lld%lld%lld%lld",&d,&k,&a,&b,&t);
LL last = d-k;
if(last<=0)printf("%lld\n",d*a);
else
{
d = last;
ans+=k*a;
LL t1 = d*b; //剩下的路程全部走路
LL t2 = (d/k)*k*a+(d/k)*t+(d-(d/k)*k)*b; //剩下的路程开车到小于k走路
LL t3 = d*a+((d/k)+1)*t; //一直开车
ans+=min(t1,min(t2,t3));
printf("%lld\n",ans);
}
}
D. Road to Post Office
time limit per test
memory limit per test
input
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
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
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
- k
- a
- b
- t
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.
