Codeforces Round #700 (Div. 2) --- B(思维)

传送门

有必要提醒自己一下, 先把题读利索了(手动捂脸)

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题目

B. The Great Hero

 

The great hero guards the country where Homer lives. The hero has attack power AA and initial health value BB. There are nn monsters in front of the hero. The ii-th monster has attack power aiai and initial health value bibi.

The hero or a monster is said to be living, if his or its health value is positive (greater than or equal to 11); and he or it is said to be dead, if his or its health value is non-positive (less than or equal to 00).

In order to protect people in the country, the hero will fight with monsters until either the hero is dead or all the monsters are dead.

• In each fight, the hero can select an arbitrary living monster and fight with it. (当时题意都理解错了, 好惨)Suppose the ii-th monster is selected, and the health values of the hero and the ii-th monster are xx and yy before the fight, respectively. After the fight, the health values of the hero and the ii-th monster become x−aix−ai and y−Ay−A, respectively.

Note that the hero can fight the same monster more than once.

For the safety of the people in the country, please tell them whether the great hero can kill all the monsters (even if the great hero himself is dead after killing the last monster).

Input

Each test contains multiple test cases. The first line contains tt (1≤t≤1051≤t≤105) — the number of test cases. Description of the test cases follows.

The first line of each test case contains three integers AA (1≤A≤1061≤A≤106), BB (1≤B≤1061≤B≤106) and nn (1≤n≤1051≤n≤105) — the attack power of the great hero, the initial health value of the great hero, and the number of monsters.

The second line of each test case contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤1061≤ai≤106), where aiai denotes the attack power of the ii-th monster.

The third line of each test case contains nn integers b1,b2,…,bnb1,b2,…,bn (1≤bi≤1061≤bi≤106), where bibi denotes the initial health value of the ii-th monster.

It is guaranteed that the sum of nn over all test cases does not exceed 105105.

Output

For each test case print the answer: "YES" (without quotes) if the great hero can kill all the monsters. Otherwise, print "NO" (without quotes).

Example

input

5
3 17 1
2
16
10 999 3
10 20 30
100 50 30
1000 1000 4
200 300 400 500
1000 1000 1000 1000
999 999 1
1000
1000
999 999 1
1000000
999

output

YES
YES
YES
NO
YES

Note

In the first example: There will be 66 fights between the hero and the only monster. After that, the monster is dead and the health value of the hero becomes 17−6×2=5>017−6×2=5>0. So the answer is "YES", and moreover, the hero is still living.

In the second example: After all monsters are dead, the health value of the hero will become 709709, regardless of the order of all fights. So the answer is "YES".

In the third example: A possible order is to fight with the 11-st, 22-nd, 33-rd and 44-th monsters. After all fights, the health value of the hero becomes −400−400. Unfortunately, the hero is dead, but all monsters are also dead. So the answer is "YES".

In the fourth example: The hero becomes dead but the monster is still living with health value 1000−999=11000−999=1. So the answer is "NO".

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题意

英雄的攻击强度为A，初始生命值为B。英雄前面有n只怪物。
下面2行, 每行n个值, 表示第i个怪物有攻击力ai和初始生命值bi。
在每场战斗中，英雄可以任意选择一个活着的怪物并与之战斗。1 vs 1
假设第i个怪物被选中，英雄和第i个怪物血量x和y --> x−ai和y−A。
注意，英雄可以多次对抗同一个怪物。
为了国家人民的安全，请告诉他们大英雄是否能杀死所有的怪物(即使大英雄杀死最后一个怪物后自己也死了)。

数据范围：A,B,a(i),b(i)<=1e6 ，n<=1e5

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思路

最后一次攻击前, 英雄血量>0, 就成功了  ===> 最后抵御最大攻击前, 英雄血量 > 0

AC代码

 1 #include <iostream>
 2 #include <algorithm>
 3
 4 using namespace std;
 5
 6 typedef long long ll;
 7
 8 const int N = 1e5 + 10;
 9
10 ll tar1[N], blood1[N], res, tar, blood;
11
12 int main()
13 {
14     int t;
15     scanf("%d", &t);
16     while(t--)
17     {
18         int n;
19         scanf("%lld%lld%d", &tar, &blood, &n);
20
21         for(int i = 0; i < n ; i++)    scanf("%lld", &tar1[i]);
22         for(int i = 0; i < n ; i++)    scanf("%lld", &blood1[i]);
23
24         res = 0;
25         ll maxn = -1;
26         for(int i = 0; i < n ; i++)
27         {
28             int sum = blood1[i] / tar + (blood1[i] % tar != 0);
29             res += sum * tar1[i];
30             maxn = max(maxn, tar1[i]);
31         }
32
33         if(res - maxn >= blood)
34             printf("NO\n");
35         else
36             printf("YES\n");
37     }
38
39     return 0;
40 }