Codeforces Round 379 (Div. 2)
C
Anton and Making Potions
Codeforces
1600
原题
← BD →
翻译中…

Anton is playing a very interesting computer game, but now he is stuck at one of the levels. To pass to the next level he has to prepare $*n*$ potions.

翻译中…

Anton has a special kettle, that can prepare one potions in $*x*$ seconds. Also, he knows spells of two types that can faster the process of preparing potions.

翻译中…
  1. Spells of this type speed up the preparation time of one potion. There are $*m*$ spells of this type, the $*i*$-th of them costs $*b*_{*i*}$ manapoints and changes the preparation time of each potion to $*a*_{*i*}$ instead of $*x*$.
  2. Spells of this type immediately prepare some number of potions. There are $*k*$ such spells, the $*i*$-th of them costs $*d*_{*i*}$ manapoints and instantly create $*c*_{*i*}$ potions.
翻译中…

Anton can use no more than one spell of the first type and no more than one spell of the second type, and the total number of manapoints spent should not exceed $*s*$. Consider that all spells are used instantly and right before Anton starts to prepare potions.

翻译中…

Anton wants to get to the next level as fast as possible, so he is interested in the minimum number of time he needs to spent in order to prepare at least $*n*$ potions.

翻译中…

Input

翻译中…

The first line of the input contains three integers $*n*$, $*m*$, $*k*$ ($1 ≤ *n* ≤ 2·10^{9}, 1 ≤ *m*, *k* ≤ 2·10^{5}$) — the number of potions, Anton has to make, the number of spells of the first type and the number of spells of the second type.

翻译中…

The second line of the input contains two integers $*x*$ and $*s*$ ($2 ≤ *x* ≤ 2·10^{9}, 1 ≤ *s* ≤ 2·10^{9}$) — the initial number of seconds required to prepare one potion and the number of manapoints Anton can use.

翻译中…

The third line contains $*m*$ integers $*a*_{*i*}$ ($1 ≤ *a*_{*i*} < *x*$) — the number of seconds it will take to prepare one potion if the $*i*$-th spell of the first type is used.

翻译中…

The fourth line contains $*m*$ integers $*b*_{*i*}$ ($1 ≤ *b*_{*i*} ≤ 2·10^{9}$) — the number of manapoints to use the $*i*$-th spell of the first type.

翻译中…

There are $*k*$ integers $*c*_{*i*}$ ($1 ≤ *c*_{*i*} ≤ *n*$) in the fifth line — the number of potions that will be immediately created if the $*i*$-th spell of the second type is used. It's guaranteed that $*c*_{*i*}$ are not decreasing, i.e. $*c*_{*i*} ≤ *c*_{*j*}$ if $*i* < *j*$.

翻译中…

The sixth line contains $*k*$ integers $*d*_{*i*}$ ($1 ≤ *d*_{*i*} ≤ 2·10^{9}$) — the number of manapoints required to use the $*i*$-th spell of the second type. It's guaranteed that $*d*_{*i*}$ are not decreasing, i.e. $*d*_{*i*} ≤ *d*_{*j*}$ if $*i* < *j*$.

翻译中…

Output

翻译中…

Print one integer — the minimum time one has to spent in order to prepare $*n*$ potions.

翻译中…

Note

翻译中…

In the first sample, the optimum answer is to use the second spell of the first type that costs $10$ manapoints. Thus, the preparation time of each potion changes to $4$ seconds. Also, Anton should use the second spell of the second type to instantly prepare $15$ potions spending $80$ manapoints. The total number of manapoints used is $10 + 80 = 90$, and the preparation time is $4·5 = 20$ seconds ($15$ potions were prepared instantly, and the remaining $5$ will take $4$ seconds each).

翻译中…

In the second sample, Anton can't use any of the spells, so he just prepares $20$ potions, spending $10$ seconds on each of them and the answer is $20·10 = 200$.

翻译中…

样例

Input #1
20 3 2
10 99
2 4 3
20 10 40
4 15
10 80
Output #1
20
InputCopy OutputCopy
Input #2
20 3 2
10 99
2 4 3
200 100 400
4 15
100 800
Output #2
200
InputCopy OutputCopy
题目信息
题号
734C
时间限制
4 s
内存限制
256 MB
标签
二分
动态规划
贪心
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。