time limit per test：1 second
memory limit per test：256 megabytes
input：standard input
output：standard output
Alice and Bob love playing one-dimensional battle ships. They play on the field in the form of a line consisting of n square cells (that is, on a 1 × n table).
At the beginning of the game Alice puts k ships on the field without telling their positions to Bob. Each ship looks as a 1 × a rectangle (that is, it occupies a sequence of a consecutive squares of the field). The ships cannot intersect and even touch each other.
After that Bob makes a sequence of "shots". He names cells of the field and Alice either says that the cell is empty ("miss"), or that the cell belongs to some ship ("hit").
But here's the problem! Alice like to cheat. May be that is why she responds to each Bob's move with a "miss".
Help Bob catch Alice cheating — find Bob's first move, such that after it you can be sure that Alice cheated.
Input
The first line of the input contains three integers: n, k and a (1 ≤ n, k, a ≤ 2·105) — the size of the field, the number of the ships and the size of each ship. It is guaranteed that the n, k and a are such that you can put k ships of size a on the field, so that no two ships intersect or touch each other.
The second line contains integer m (1 ≤ m ≤ n) — the number of Bob's moves.
The third line contains m distinct integers x1, x2, ..., xm, where xi is the number of the cell where Bob made the i-th shot. The cells are numbered from left to right from 1 to n.
Output
Print a single integer — the number of such Bob's first move, after which you can be sure that Alice lied. Bob's moves are numbered from 1 to m in the order the were made. If the sought move doesn't exist, then print "-1".
Examples
Input
11 3 3 5 4 8 6 1 11
Output
3
Input
5 1 3 2 1 5
Output
-1
Input
5 1 3 1 3
Output
1
題意：在n*1的網格中的可以保證能夠放置k個長度爲a的船，Bob對船進行射擊，則表示被射擊的位置不可以放船，求Bob射擊移動到第幾個位置的時候n*1的網格放不下k個小船，注意：船與船之間不相鄰，至少必須空一格。由這一條信息可以知道一艘船佔地面積爲a+1，最初始時n*1的網格中最後放置的一艘船佔地面積爲a，不算空格，所以我們需要模擬編號爲n+1的網格用來存放空格，所以最初始時n*1的網格中最多可以放（n+1）/（a+1）艘船。然後每射擊一次計算當前網格中最多可以按要求放置多少條船，用二分法計算，最後若放置的船小於題目要求的數量則輸出射擊的次數。
就例1來看：（隨手畫的有點小丑）
兩個穿之間不相鄰這點一定要注意，我用了各種奇奇怪怪的方法寫寫了我一兩個小時最後還是因爲這一點WA了，/(ㄒoㄒ)/~~
#include <iostream>
#include <set>
using namespace std;
int main()
{
set<int>s;
int n,k,a,m,d;
cin>>n>>k>>a>>m;
s.insert(0);
s.insert(n+1);
int ans=(n+1)/(a+1);
int cont=-1,b=0;
for(int i=1;i<=m;i++)
{
cin>>d;
int e=*s.lower_bound(d);
int f=*--s.lower_bound(d);
s.insert(d)
ans-=(e-f)/(a+1);
ans+=(e-d)/(a+1)+(d-f)/(a+1);
if(ans<k&&b==0){cont=i;b=1;}
}
cout<<cont<<endl;
}