3856: Monster

最新推荐文章于 2018-11-12 21:28:00 发布

转载最新推荐文章于 2018-11-12 21:28:00 发布 · 69 阅读

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原文链接：http://www.cnblogs.com/HansBug/p/4486524.html

本文解析了3856:Monster题目，介绍了TeacherMai如何通过攻击来减少怪物的生命值，并判断是否能在休息前或经过若干轮攻击后使怪物生命值降至1以下，最终给出解答。

3856: Monster

Time Limit: 1 Sec Memory Limit: 64 MB
Submit: 351 Solved: 161
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Description

Teacher Mai has a kingdom. A monster has invaded this kingdom, and Teacher Mai wants to kill it.

Monster initially has h HP. And it will die if HP is less than 1.

Teacher Mai and monster take turns to do their action. In one round, Teacher Mai can attack the monster so that the HP of the monster will be reduced by a. At the end of this round, the HP of monster will be increased by b.

After k consecutive round's attack, Teacher Mai must take a rest in this round. However, he can also choose to take a rest in any round.

Output "YES" if Teacher Mai can kill this monster, else output "NO".

Input

There are multiple test cases, terminated by a line "0 0 0 0".

For each test case, the first line contains four integers h,a,b,k(1<=h,a,b,k <=10^9).

Output

For each case, output "Case #k: " first, where k is the case number counting from 1. Then output "YES" if Teacher Mai can kill this monster, else output "NO".

Sample Input

5 3 2 2
0 0 0 0

Sample Output

Case #1: NO

HINT

Source

By 镇海中学

题解：其实，这是一道水题——事实上，的确是道水题——但是，但是——很明显此题很容易WA掉= =

想干掉这个怪，必须要满足三个条件之一——1.可以一击必杀 2.在休息之前可以干掉这个怪 3.每次进行了一个K长度的周期后这个怪的总体血量在下降

（PS：想起来满是童年口袋妖怪的影子啊，但我居然还是WA掉了2次= =）

 1 /**************************************************************
 2     Problem: 3856
 3     User: HansBug
 4     Language: Pascal
 5     Result: Accepted
 6     Time:4 ms
 7     Memory:224 kb
 8 ****************************************************************/
 9  
10 var
11    i,j,k,l,m,n:int64;
12 begin
13      while not(eof) do
14            begin
15                 readln(i,j,k,l);inc(m);
16                 if (i=0) and (j=0) and (k=0) and (l=0) then exit;
17                 write('Case #',m,': ');
18                 if (j>=i) or ((l*j-(l-1)*k)>=i) or ((l*j)>((l+1)*k)) then writeln('YES') else writeln('NO');
19            end;
20 end.