CF 567D (One-dimen1_battle Ships-binary), battleships

CF 567D (One-dimen1_battle Ships-binary), battleships

D. One-dimen1_battle Shipstime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard output

Alice and Bob love playing one-dimen1_battle ships. They play on the field in the form of a line consistingNSquare cells (that is, on a 1 hour × hourNTable ).

At the beginning of the game Alice putsKShips on the field without telling their positions to Bob. Each ship looks as a 1 minute × hourARectangle (that is, it occupies a sequenceAConsecutive 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,KAndA(1 digit ≤ DigitN, Bytes,K, Bytes,ALimit ≤ limit 2. 105)-the size of the field, the number of the ships and the size of each ship. It is guaranteed thatN,KAndAAre such that you can putKShips of sizeAOn the field, so that no two ships intersect or touch each other.

The second line contains integerM(1 digit ≤ DigitMLimit ≤ limitN)-The number of Bob's moves.

The third line containsMDistinct integersX1, bytes,X2, middle..., middle ,...,XM, WhereXIIs the number of the cell where Bob madeI-Th shot. The cells are numbered from left to right from 1N.

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 from1MIn the order the were made. If the sought move doesn't exist, then print "-1 ".

Sample test (s) input

11 3 354 8 6 1 11

Output

3

Input

5 1 321 5

Output

-1

Input

5 1 313

Output

1

Bare binary

#include<cstdio>#include<cstring>#include<cstdlib>#include<algorithm>#include<functional>#include<iostream>#include<cmath>#include<cctype>#include<ctime>using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define Forp(x) for(int p=pre[x];p;p=next[p])#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  #define Lson (x<<1)#define Rson ((x<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,127,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define INF (2139062143)#define F (100000007)#define MAXN (200000+10)typedef long long ll;ll mul(ll a,ll b){return (a*b)%F;}ll add(ll a,ll b){return (a+b)%F;}ll sub(ll a,ll b){return (a-b+(a-b)/F*F+F)%F;}void upd(ll &a,ll b){a=(a%F+b%F)%F;}int n,k,a,m;int x[MAXN],x2[MAXN];bool check(int m){For(i,m) x2[i]=x[i];sort(x2+1,x2+1+m);int l=1,tot=0;For(i,m) {int len=x2[i]-l+1;tot+=len/(a+1);l=x2[i]+1;}tot+=(n+1-l+1)/(a+1);if (tot>=k) return 1;return 0;}int main(){//freopen("D.in","r",stdin);//freopen(".out","w",stdout);cin>>n>>k>>a>>m;For(i,m) scanf("%d",&x[i]);int l=1,r=m,ans=INF;while(l<=r){int m=(r+l)/2;if (check(m)) l=m+1;else r=m-1,ans=min(ans,m);}if (ans==INF) ans=-1;cout<<ans<<endl;return 0;}

Copyright Disclaimer: This article is an original article by the blogger and cannot be reproduced without the permission of the blogger.