[Codeforces] #436 C. Bus-优快云博客

本文介绍了一个关于公交车加油策略的问题，公交车需要在Ox坐标线上往返行驶，每次从0到a再返回算作一次旅程，途中只有一个加油站位于f点，讨论如何在有限油量下完成k次旅程所需的最少加油次数。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the pointx = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0is called a bus journey. In total, the bus must make k journeys.

The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.

There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.

What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the pointx = 0.

Input

The first line contains four integers a, b, f, k (0 < f < a ≤ 106, 1 ≤ b ≤ 109, 1 ≤ k ≤ 104) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.

Output

Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print-1.

Examples

input

6 9 2 4

output

input

6 10 2 4

output

input

6 5 4 3

output

-1

Note

In the first example the bus needs to refuel during each journey.

In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.

In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.

Analysis

公交车= =

存在这样一条数轴：被称为Ox

（所以我们知道坐标系可能被叫做Oxy）

公交车要干的事情，就是从 0 跑到 a

然后再跑回来

根据题意 “Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey.”

我们知道单程就是一次Journey

公交车总共要跑 k 次单程

而在 f 有一个油井（假装是油井 =w= 才够霸气）

公交车可以选择加满油或者不加

加满就是 b 升了

由于只有这一个油井，所以如果公交车需要预判一下未来的形势，决定要不要加油

最后输出公交车需要加油的最少次数（节约能源 =w= )

-------------------------------------------------------

由于数据比较小，可以直接暴力模拟

如果公交车还需要走回来的话，在油井的时候需要预判剩下的油够不够走回油井，否则预判能否到达终点即可

=w= 越简单写起来越容易错，，

Code

 1 #include<cstdio>
 2 #include<iostream>
 3 using namespace std;
 4 
 5 int a,b,f,k,ans,tank;
 6 
 7 int main(){
 8     scanf("%d%d%d%d",&a,&b,&f,&k);
 9     
10     tank = b;
11     
12     for(int i = 1;i <= k;i++){
13         if(i%2){ //0 -> a
14             if(tank < f){ printf("-1"); return 0; }
15             
16             tank -= f;
17             
18             // Refruel
19             if(tank < ((i < k)?(2*(a-f)):(a-f))){
20 //                printf("#%d: %d\n",k,tank);
21                 tank = b,ans++;
22             }
23             
24             
25             
26             // Can't arr
27             if(tank < (a-f)){ printf("-1"); return 0; }
28             
29             tank -= (a-f);
30         }else{   //a -> 0
31             // Can't refruel?
32             if(tank < (a-f)){ printf("-1"); return 0; }
33 //            
34             tank -= (a-f);
35             
36             if(tank < ((i < k)?(2*f):(f))){
37 //                printf("#%d: %d\n",k,tank);
38                 tank = b,ans++;
39             }
40             
41             if(tank < f){ printf("-1"); return 0; }
42             
43             tank -= f;
44         }
45     }
46     
47     printf("%d",ans);
48     
49     return 0;
50 }