B. The Modcrab
Vova is again playing some computer game, now an RPG. In the game Vova‘s character received a quest: to slay the fearsome monster called Modcrab.
After two hours of playing the game Vova has tracked the monster and analyzed its tactics. The Modcrab has h2 health points and an attack power of a2. Knowing that, Vova has decided to buy a lot of strong healing potions and to prepare for battle.
Vova‘s character has h1 health points and an attack power of a1. Also he has a large supply of healing potions, each of which increases his current amount of health points by c1 when Vova drinks a potion. All potions are identical to each other. It is guaranteed that c1 > a2.
The battle consists of multiple phases. In the beginning of each phase, Vova can either attack the monster (thus reducing its health by a1) or drink a healing potion (it increases Vova‘s health by c1; Vova‘s health can exceed h1). Then, if the battle is not over yet, the Modcrab attacks Vova, reducing his health by a2. The battle ends when Vova‘s (or Modcrab‘s) health drops to 0 or lower. It is possible that the battle ends in a middle of a phase after Vova‘s attack.
Of course, Vova wants to win the fight. But also he wants to do it as fast as possible. So he wants to make up a strategy that will allow him to win the fight after the minimum possible number of phases.
Help Vova to make up a strategy! You may assume that Vova never runs out of healing potions, and that he can always win.
Input
The first line contains three integers h1, a1, c1 (1 ≤ h1, a1 ≤ 100, 2 ≤ c1 ≤ 100) — Vova‘s health, Vova‘s attack power and the healing power of a potion.
The second line contains two integers h2, a2 (1 ≤ h2 ≤ 100, 1 ≤ a2 < c1) — the Modcrab‘s health and his attack power.
Output
In the first line print one integer n denoting the minimum number of phases required to win the battle.
Then print n lines. i-th line must be equal to HEAL if Vova drinks a potion in i-th phase, or STRIKE if he attacks the Modcrab.
The strategy must be valid: Vova‘s character must not be defeated before slaying the Modcrab, and the monster‘s health must be 0 or lower after Vova‘s last action.
If there are multiple optimal solutions, print any of them.
Input
10 6 100 17 5
Output
4 STRIKE HEAL STRIKE STRIKE
Input
11 6 100 12 5
Output
2 STRIKE STRIKE
题意：vova要打败一个怪物，vova 的血量h1,攻击力a1，可以无限使用的血瓶c1，可以加c1这么多血，加的血量可以无上限。怪物的血量h2，攻击力a2。
思路：模拟就行。【注意】一开始没考虑，可以一击必杀的那种情况，就是h2怪物血量小于，vova攻击力并且vova血量小于怪物攻击力，其实不用加血的，直接秒杀
AC代码：
1 #include<bits/stdc++.h>
2
3 using namespace std;
4
5 #define int long long
6 int arr[100500];
7 signed main(){
8 int h1,a1,c1;
9 int h2,a2;
10 cin>>h1>>a1>>c1;
11 cin>>h2>>a2;
12 if(a1>=h2){
13 printf("1\nSTRIKE");
14 return 0;
15 }
16 int x1=0;
17 int x2=0;
18 if(h2%a1){
19 x1++;
20 }
21 if(h1%a2){
22 x2++;
23 }
24 x1+=h2/a1;
25 x2+=h1/a2;
26 if(x2>=x1){
27 printf("%d\n",x1);
28 for(int i=0;i<x1;i++){
29 printf("STRIKE\n");
30 }
31 return 0;
32 }
33 int cnt=1;
34 while(1){
35 if((h2-a1<=0)){// 【注意】：可以一击必杀的QAQ
36 arr[cnt++]=1;
37 break;
38 }
39 if((h1-a2)<=0){
40 arr[cnt++]=0;
41 h1+=c1;
42 }else{
43 h2=h2-a1;
44 arr[cnt++]=1;
45 if(h2<=0){
46 break;
47 }
48
49 }
50 h1-=a2;
51 }
52 printf("%d\n",cnt-1);
53 for(int i=1;i<cnt;i++){
54 if(arr[i]==1){
55 printf("STRIKE\n");
56 }else{
57 printf("HEAL\n");
58 }
59 }
60 return 0;
61 }
原文地址：https://www.cnblogs.com/pengge666/p/11552476.html