Codeforces Round 610 (Div. 2)
C
Petya and Exam
Codeforces
1800
原题
← B2D →
翻译中…

Petya has come to the math exam and wants to solve as many problems as possible. He prepared and carefully studied the rules by which the exam passes.

翻译中…

The exam consists of $n$ problems that can be solved in $T$ minutes. Thus, the exam begins at time $0$ and ends at time $T$. Petya can leave the exam at any integer time from $0$ to $T$, inclusive.

翻译中…

All problems are divided into two types:

翻译中…
  • easy problems — Petya takes exactly $a$ minutes to solve any easy problem;
  • hard problems — Petya takes exactly $b$ minutes ($b \gt a$) to solve any hard problem.
翻译中…

Thus, if Petya starts solving an easy problem at time $x$, then it will be solved at time $x+a$. Similarly, if at a time $x$ Petya starts to solve a hard problem, then it will be solved at time $x+b$.

翻译中…

For every problem, Petya knows if it is easy or hard. Also, for each problem is determined time $t_i$ ($0 \le t_i \le T$) at which it will become mandatory (required). If Petya leaves the exam at time $s$ and there is such a problem $i$ that $t_i \le s$ and he didn't solve it, then he will receive $0$ points for the whole exam. Otherwise (i.e if he has solved all such problems for which $t_i \le s$) he will receive a number of points equal to the number of solved problems. Note that leaving at time $s$ Petya can have both "mandatory" and "non-mandatory" problems solved.

翻译中…

For example, if $n=2$, $T=5$, $a=2$, $b=3$, the first problem is hard and $t_1=3$ and the second problem is easy and $t_2=2$. Then:

翻译中…
  • if he leaves at time $s=0$, then he will receive $0$ points since he will not have time to solve any problems;
  • if he leaves at time $s=1$, he will receive $0$ points since he will not have time to solve any problems;
  • if he leaves at time $s=2$, then he can get a $1$ point by solving the problem with the number $2$ (it must be solved in the range from $0$ to $2$);
  • if he leaves at time $s=3$, then he will receive $0$ points since at this moment both problems will be mandatory, but he will not be able to solve both of them;
  • if he leaves at time $s=4$, then he will receive $0$ points since at this moment both problems will be mandatory, but he will not be able to solve both of them;
  • if he leaves at time $s=5$, then he can get $2$ points by solving all problems.
翻译中…

Thus, the answer to this test is $2$.

翻译中…

Help Petya to determine the maximal number of points that he can receive, before leaving the exam.

翻译中…

Input

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The first line contains the integer $m$ ($1 \le m \le 10^4$) — the number of test cases in the test.

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The next lines contain a description of $m$ test cases.

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The first line of each test case contains four integers $n, T, a, b$ ($2 \le n \le 2\cdot10^5$, $1 \le T \le 10^9$, $1 \le a \lt b \le 10^9$) — the number of problems, minutes given for the exam and the time to solve an easy and hard problem, respectively.

翻译中…

The second line of each test case contains $n$ numbers $0$ or $1$, separated by single space: the $i$-th number means the type of the $i$-th problem. A value of $0$ means that the problem is easy, and a value of $1$ that the problem is hard.

翻译中…

The third line of each test case contains $n$ integers $t_i$ ($0 \le t_i \le T$), where the $i$-th number means the time at which the $i$-th problem will become mandatory.

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It is guaranteed that the sum of $n$ for all test cases does not exceed $2\cdot10^5$.

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Output

翻译中…

Print the answers to $m$ test cases. For each set, print a single integer — maximal number of points that he can receive, before leaving the exam.

翻译中…

样例

Input #1
10
3 5 1 3
0 0 1
2 1 4
2 5 2 3
1 0
3 2
1 20 2 4
0
16
6 20 2 5
1 1 0 1 0 0
0 8 2 9 11 6
4 16 3 6
1 0 1 1
8 3 5 6
6 20 3 6
0 1 0 0 1 0
20 11 3 20 16 17
7 17 1 6
1 1 0 1 0 0 0
1 7 0 11 10 15 10
6 17 2 6
0 0 1 0 0 1
7 6 3 7 10 12
5 17 2 5
1 1 1 1 0
17 11 10 6 4
1 1 1 2
0
1
Output #1
3
2
1
0
1
4
0
1
2
1
InputCopy OutputCopy
题目信息
题号
1282C
时间限制
2 s
内存限制
256 MB
标签
贪心
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双指针

题面与样例版权归 Codeforces 及出题人所有;中文翻译由 AI 生成(CI 管线),如有出入请以原文为准。