Codeforces Round #Pi (Div. 2) D. One-Dimensional Battle Ships set乱搞

D. One-Dimensional Battle Ships
Time Limit: 2 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/567/problem/D

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 ≤ 2·105) — 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 from1 to m in the order the were made. If the sought move doesn't exist, then print "-1".

Sample Input

11 3 3
5
4 8 6 1 11

Sample Output

3

HINT

题意

   在一维区间上，每次击打一个长度为1的区间，问在第几次（<=m）不能放这k*a的船，注意船与船不相互接触；

   不存在输出-1；

题解：

    set寻找相邻x端点 维护值判断下就好了

代码

#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <queue>
#include <typeinfo>
#include <map>
#include <stack>
typedef __int64 ll;
#define inf 1000000000000
using namespace std;
inline ll read()
{
    ll x=0,f=1;
    char ch=getchar();
    while(ch<'0'||ch>'9')
    {
        if(ch=='-')f=-1;
        ch=getchar();
    }
    while(ch>='0'&&ch<='9')
    {
        x=x*10+ch-'0';
        ch=getchar();
    }
    return x*f;
}

//**************************************************************************************

set<int >s;
set<int >::iterator it,itt;
int n,k,a;
int sum;
bool test(int x)
{
    it=s.lower_bound(x);
    int r=*it;
    int l=*(--it);
    sum-=(r-l)/(a+1);
    sum+=((x-l)/(a+1)+(r-x)/(a+1));
    if(sum<k)return true;
    else return false;
}
int main()
{
    scanf("%d%d%d",&n,&k,&a);
    int m;
    scanf("%d",&m);
    s.insert(0);
    s.insert(n+1);
    int x;
    sum=(n+1)/(a+1);
    if(sum<k)
    {
        printf("-1
");
        return 0;
    }
    for(int i=1; i<=m; i++)
    {
        scanf("%d",&x);
        if(s.count(x))continue;
        // for(it=s.begin();it!=s.end();it++)printf("%d
",*it);
        if(test(x))
        {
            printf("%d
",i);
            return 0;
        }//printf("%d
",sum);
        s.insert(x);
    }
    printf("-1
");

    return 0;
}