从头到尾枚举一遍就行
#include <bits/stdc++.h>
#include <mutex>
#define int long long
using namespace std;
#define endl '\n'
#define pb push_back
#define ull unsigned long long
#define all(a) a.begin(), a.end()
#define vi vector<int>
#define vii vector<vector<int>>
#define fi first
#define se second
#define vs vector<string>
#define eb emplace_back
#define in insert
#define pf push_front
#define rep(i, n) for (int i = 0; i < (n); ++i)
#define rep1(i, n) for (int i = 1; i <= (n); i++)
const int inf = 2e18 + 9 ;
const int mod1 = 1e9 + 7 ;
const int mod2 = 998244353 ;
typedef pair<int, int> pll;
typedef long double db;
inline int gcd(int a , int b) {return b ? gcd(b , a % b) : a;};
template<class T>inline void read(T &res){char c;T flag=1;while((c=getchar())<'0'||c>'9')if(c=='-')flag=-1;res=c-'0';while((c=getchar())>='0'&&c<='9')res=res*10+c-'0';res*=flag;}
inline int qsm(int a , int b){int res = 1 ; while(b){if(b & 1){res *= a ; }b >>= 1 ; a *= a ; }return res ; }
int dir[4][2] = {{1 , 0} , {-1 , 0} , {0 , 1} , {0 , -1}};
int dirx[8] = {-1, -1, -1, 0, 0, 1, 1, 1};
int diry[8] = {-1, 0, 1, -1, 1, -1, 0, 1};
struct ti
{
int time , nan ;
};
void work()
{
int n , t ; cin >> n >> t ;
int a , b ; cin >> a >> b ;
int a1 = 0 ; int b1 = 0 ;
int at = t ; int bt = t ;
vector<ti>arr(n);
for(int i = 0 ; i < n ; i++)
{
cin >> arr[i].time ;
}
for(int i = 0 ; i < n ; i++)
{
cin >> arr[i].nan ;
}
for(int i = 0 ; i < n ; i++)
{
if(arr[i].nan >= b && bt >= 2 * arr[i].time)
{
b1++ ;
bt -= 2 * arr[i].time ;
}
else if(arr[i].nan < b && bt >= arr[i].time)
{
bt -= arr[i].time ;
b1++ ;
}
if(arr[i].nan < a && at >= arr[i].time)
{
a1++ ;
at -= arr[i].time ;
}
}
cout << a1 << " " << b1 << endl ;
}
signed main()
{
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int t = 1;
while (t--)
{
work();
}
return 0;
}

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