Codeforces 567D：One-Dimensional Battle Ships（二分）

time limit per test : 1 second
memory limit per test : 256 megabytes
input : standard input
output : standard output

Problem Description

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 \(a (1 ≤ n, k, a ≤ 2·10^5)\) — 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 \(x_1, x_2, ..., x_m\), where \(x_i\) 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

题意

一个长为\(1\times n\)的区域内有\(k\)个战舰，每个战舰的长度均为\(a\)，Bob射击\(m\)个不同的位置，对于每次射击，Alice都会告诉Bob是否击中。

但是因为Alice会撒谎，所以当Bob击中的时候，Alice也会说没有击中

问Bob最早可以在第几次射击的时候发现Alice在说谎

思路

因为随着射击次数的增多，Bob发现Alice说谎的可能性越大，所以可以利用二分来解决

将\([1,mid]\)区间的位置进行排序，然后计算这些位置如果全部没有击中，可以放下多少战舰，如果超过\(k\)个，那么当前的\(mid\)是可行的，查询前半部分，否则，查询后半部分

代码

#include <bits/stdc++.h>
#define ll long long
#define ull unsigned long long
#define ms(a,b) memset(a,b,sizeof(a))
const int inf=0x3f3f3f3f;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int maxn=1e6+10;
const int mod=1e9+7;
const int maxm=1e3+10;
using namespace std;
int x[maxn];
int b[maxn];
int n,k,a;
bool check(int mid)
{
	for(int i=1;i<=mid;i++)
		b[i]=x[i];
	sort(b+1,b+1+mid);
	int cnt=0;
	for(int i=1;i<=mid;i++)
		cnt+=(b[i]-b[i-1])/(a+1);	// b[i]和b[i-1]之间可以放多少战舰
	// b[mid]->最后一个位置可以放多少战舰
	cnt+=(n-b[mid]+1)/(a+1);
	if(cnt>=k)
		return true;
	return false;
}
int main(int argc, char const *argv[])
{
    #ifndef ONLINE_JUDGE
        freopen("/home/wzy/in", "r", stdin);
        freopen("/home/wzy/out", "w", stdout);
        srand((unsigned int)time(NULL));
    #endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin>>n>>k>>a;
    int m;
    cin>>m;
    for(int i=1;i<=m;i++)
    {
    	cin>>x[i];
    }
	int l=1,r=m,ans=inf;
	while(l<=r)
	{
		int mid=(l+r)/2;
		if(check(mid))
			l=mid+1;
		else
			r=mid-1,ans=min(ans,mid);
	}
	if(ans==inf)
		cout<<-1<<endl;
	else
		cout<<ans<<endl;
    #ifndef ONLINE_JUDGE
        cerr<<"Time elapsed: "<<1.0*clock()/CLOCKS_PER_SEC<<" s."<<endl;
    #endif
    return 0;
}