Codeforces 702 D Road to Post Office

题目描述

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

Vasiliy‘s car is not new — it breaks after driven every kk kilometers and Vasiliy needs tt seconds to repair it. After repairing his car Vasiliy can drive again (but after kk 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 aa seconds, to walk one kilometer on foot he needs bb seconds ( a<ba<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.

输入输出格式

输入格式：

The first line contains 5 positive integers d,k,a,b,td,k,a,b,t ( 1<=d<=10^{12}1<=d<=1012 ; 1<=k,a,b,t<=10^{6}1<=k,a,b,t<=106 ; a<ba<b ), where:

• dd — the distance from home to the post office;

• kk — the distance, which car is able to drive before breaking;

• aa — the time, which Vasiliy spends to drive 1 kilometer on his car;

• bb — the time, which Vasiliy spends to walk 1 kilometer on foot;

• tt — the time, which Vasiliy spends to repair his car.

输出格式：

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

输入输出样例

5 2 1 4 10

14

5 2 1 4 5

13

说明

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.

小学数学题。。。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<vector>
#define ll long long
using namespace std;
ll cost,d,b,a,k,t,n;
int main(){
    scanf("%lld%lld%lld%lld%lld",&d,&k,&a,&b,&t);
    ll xl=(a-b)*k+t,tim=d/k+(d%k?1:0);
    cost=min(b*d,d*a+(tim-1)*t);

    if(k<=d){
        if(xl>0) cost=min(cost,b*d-t+xl);
        else{
            n=d/k;
            cost=min(cost,xl*n+b*d-t);
        }
    }
    
    printf("%lld\n",cost);
    return 0;
}