AtCoder-----Katana Thrower

Problem Statement

You are going out for a walk, when you suddenly encounter a monster. Fortunately, you have N katana (swords), Katana 1, Katana 2, …, Katana N, and can perform the following two kinds of attacks in any order:

• Wield one of the katana you have. When you wield Katana i (1≤i≤N), the monster receives ai points of damage. The same katana can be wielded any number of times.
• Throw one of the katana you have. When you throw Katana i (1≤i≤N) at the monster, it receives bi points of damage, and you lose the katana. That is, you can no longer wield or throw that katana.

The monster will vanish when the total damage it has received is H points or more. At least how many attacks do you need in order to vanish it in total?

Constraints

• 1≤N≤105
• 1≤H≤109
• 1≤ai≤bi≤109
• All input values are integers.

Sample Input 1

1 10
3 5

Sample Output 1

3

You have one katana. Wielding it deals 3 points of damage, and throwing it deals 5 points of damage. By wielding it twice and then throwing it, you will deal 3+3+5=11 points of damage in a total of three attacks, vanishing the monster.

Sample Input 2

2 10
3 5
2 6

Sample Output 2

2

In addition to the katana above, you also have another katana. Wielding it deals 2 points of damage, and throwing it deals 6 points of damage. By throwing both katana, you will deal 5+6=11 points of damage in two attacks, vanishing the monster.

Sample Input 3

4 1000000000
1 1
1 10000000
1 30000000
1 99999999

Sample Output 3

860000004

Sample Input 4

5 500
35 44
28 83
46 62
31 79
40 43

Sample Output 4

9

 　　这题看起来是动归，但其实不然，因为数据范围比较大%>_<%，而且伤害还可以大于H。所以这题用贪心，首先考虑一下，题目说扔完刀之后刀就没了，就没法看了，但考虑一下这是否有影响，显然如果我扔之前砍2下，扔之后砍3下是等价于扔之前砍5下的，所以只要将伤害(不是刀)从大到小排序，排完后扫一遍，如果是飞出去的就计入伤害，直到碰到第一个刀砍的伤害，然后用这个伤害狂砍QWQ

上代码:

 1 #include <iostream>
 2 #include <algorithm>
 3 #include <cmath>
 4 #include <climits>
 5  
 6 using namespace std;
 7  
 8 struct S{
 9     int p;
10     bool isFly;
11 } a[200005];
12 int n, h;
13  
14 bool cmp(S a, S b) {
15     return a.p > b.p;
16 }
17  
18 int main() {
19     cin >> n >> h;
20     for (int i = 1; i <= n; i ++) {
21         cin >> a[i].p >> a[n + i].p;
22         a[n + i].isFly = true;
23     }
24     sort(a + 1, a + 2 * n + 1, cmp);
25     int s = 0;
26     int t = 0;
27     for (int i = 1; i <= 2 * n; i ++) {
28         if (a[i].isFly == true) {
29             t ++;
30             s += a[i].p;
31         } else {;
32             t += ceil((double)(h - s) / a[i].p);
33             s = INT_MAX;
34         }
35         if (s >= h) {
36             cout << t << endl;
37             break;
38         }
39     }
40  
41     return 0;
42 }