There are n bonuses on a line: the i-th bonus is located at the point xi. All coordinates of all bonuses are distinct. You are located at the point with the coordinate 0. How many bonuses can you collect in t seconds, if you can pass the distance 1 in one second?
Input
The first line contains two integers n and t (1≤n≤200000,0≤t≤109) — the number of bonuses and the time limit.
The second line contains n integers x1, x2,…, xn (−109≤xi≤109) — the coordinates of the bonuses. They are sorted in increasing order (x1<x2<…<xn).
Output
Output one integer — the maximum number of bonuses that can be collected in t seconds.
Example
Input
5 6
-4 -1 2 3 7
Output
3
Note
To collect 3 bonuses in t=6 seconds, you must first collect the bonus with the coordinate -1, then the bonus with the coordinate 2, and in the end — the bonus with the coordinate 3. It will take you 5 seconds (1 second to collect the bonus with the coordinate -1, and 4 seconds to go from -1 to 3).
It is impossible to come to the bonus with the coordinate 7 in time. And if you first go to the bonus with the coordinate -4, you can collect only two bonuses (-4 and -1).
| 题意：从原点开始，只能走t个单位，问能拿到的最大贡献 |
思路：
其实也就四种情况：
1.只往左走。
2.只往右走。
3.先走左边到i点，看能去到右边的哪个点。
4.先走右边到i点，看能去到左边的哪个点。
前两个情况线性取个最值就好。后两种用二分法，枚举到左边的i点，减去2*dis，dis为走到i要的费用，看剩下的能到右边哪里，一个upper_bound即可。后者同理。
AC代码：
#include<iostream>
#include<string>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<map>
#include <queue>
#include<sstream>
#include <stack>
#include <set>
#include <bitset>
#include<vector>
#define FAST ios::sync_with_stdio(false)
#define abs(a) ((a)>=0?(a):-(a))
#define sz(x) ((int)(x).size())
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
#define rep(i,a,n) for(int i=a;i<=n;++i)
#define per(i,n,a) for(int i=n;i>=a;--i)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
using namespace std;
typedef long long ll;
typedef pair<ll,ll> PII;
const int maxn = 2e5+200;
const int inf=0x3f3f3f3f;
const double eps = 1e-7;
const double pi=acos(-1.0);
const int mod = 1e9+7;
inline int lowbit(int x){return x&(-x);}
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){if(!b){d=a,x=1,y=0;}else{ex_gcd(b,a%b,d,y,x);y-=x*(a/b);}}//x=(x%(b/d)+(b/d))%(b/d);
inline ll qpow(ll a,ll b,ll MOD=mod){ll res=1;a%=MOD;while(b>0){if(b&1)res=res*a%MOD;a=a*a%MOD;b>>=1;}return res;}
inline ll inv(ll x,ll p){return qpow(x,p-2,p);}
inline ll Jos(ll n,ll k,ll s=1){ll res=0;rep(i,1,n+1) res=(res+k)%i;return (res+s)%n;}
inline ll read(){ ll f = 1; ll x = 0;char ch = getchar();while(ch>'9'|ch<'0') {if(ch=='-') f=-1; ch = getchar();}while(ch>='0'&&ch<='9') x = (x<<3) + (x<<1) + ch - '0', ch = getchar();return x*f; }
int dir[4][2] = { {1,0}, {-1,0},{0,1},{0,-1} };
ll pos[maxn];
ll neg[maxn];
ll pre1[maxn];
ll pre2[maxn];
ll p1 = 0;
ll p2 = 0;
int main()
{
ll n = read(), t = read();
ll flag = 0;
rep(i,1,n)
{
ll x = read();
if(x < 0) neg[++p1] = -x;
else if(x>0) pos[++p2] = x;
else flag = 1;
}
sort(neg+1,neg+1+p1);
pre1[0] = pre2[0] = 0;
rep(i,1,p1) pre1[i] = pre1[i-1] + neg[i];
rep(i,1,p2) pre2[i] = pre2[i-1] + pos[i];
ll ans = 0;
ll cur = 0;
rep(i,1,p1)
{
if(neg[i] > t) break;
cur ++;
}
ans = max(ans, cur) ;
cur = 0;
rep(i,1,p2)
{
if(pos[i] > t) break;
cur++;
}
ans = max(ans,cur);
rep(i,1,p1)
{
ll tmp = neg[i];
ll Left = t - neg[i] * 2;
if(Left < 0 ) break;
ll idx = upper_bound(pos+1,pos+1+p2, Left) - pos;
idx --;
cur = idx + i;
ans = max(cur,ans);
}
rep(i,1,p2)
{
ll tmp = pos[i];
ll Left = t - pos[i]*2;
if(Left < 0) break;
ll idx = upper_bound(neg+1,neg+1+p1,Left) - neg;
idx --;
cur = idx + i;
ans = max(cur,ans);
}
cout<<ans + flag<<'\n';
return 0;
}