题干：
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 point x = 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 = 0 is 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 point x = 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
4
Input
6 10 2 4
Output
2
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 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
题目大意：
输入a,b,f,k。（数据范围如题干，其实这题数据范围还是蛮重要的，，不然我也不敢这么构造了）首先定义说从坐标0~a和a~0均算是一次行程。有一辆车满载是b升油，一升油可以跑一个单位长度。中间在下标为f这个固定的点有加油站，每次加满油。出发前加满油，问你跑完k次行程最少需要加多少次油。如果跑不下来，输出-1。
解题报告：
这题代码量其实不大，，我把很多代码都花在了排除特殊情况上。。。
思路很多，，可以直接模拟，，但是代码量很大，，我是把他撸成了一条直线，，然后就成了贪心加油站问题了。。。就很简单了，，，但是特殊情况第一次提交的时候忘了排除了，，，刚开始判断输出-1的条件写错了。
AC代码：
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define ll long long
#define pb push_back
#define pm make_pair
#define fi first
#define se second
using namespace std;
const int MAX = 2e5 + 5;
ll a,b,f,k,tot;
ll dis[MAX],cur;
int main()
{
ll ans = 0;
cin>>a>>b>>f>>k;
if(k>2) {
if(b < 2*f || b < 2*(a-f)) {
puts("-1"); return 0 ;
}
}
else if(k == 2){
if(b < f || b < 2*(a-f)) {
puts("-1"); return 0 ;
}
}
else {
if(b < f || b < (a-f)) {
puts("-1"); return 0 ;
}
}
if(k&1) {
int all = k/2;
cur = 0;
for(int i = 1; i<=all; i++) {
dis[++tot] = cur+f;
dis[++tot] = cur+2*a-f;
cur += 2*a;
}
dis[++tot] = cur+f;
cur += a;
// for(int i = 1; i<=tot; i++) printf("%lld\n",dis[i]);
}
else {
int all = k/2;
cur = 0;
for(int i = 1; i<=all; i++) {
dis[++tot] = cur+f;
dis[++tot] = cur+2*a-f;
cur += 2*a;
}
// for(int i = 1; i<=tot; i++) printf("%lld\n",dis[i]);
}
dis[tot+1] = 2*a+(ll)1e10;
ll now = b,cnt = 0;
for(int i = 1; i<=tot; i++) {
if(now >= cur) break;
if(dis[i] <= now && dis[i+1] > now) {
now = dis[i] + b;
cnt++;
}
}
printf("%lld\n",cnt);
return 0 ;
}