| 4950 | Monster |
MonsterTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Problem Description Teacher Mai has a kingdom. A monster has invaded this Monster initially has h HP. Teacher Mai and monster take turns After k consecutive round‘s attack, Output "YES" if Teacher Mai can kill this monster, Input There are multiple test cases, terminated by a line "0 For each test case, the first line contains four integers Output For each case, output "Case #k: " first, where k is the Sample Input 5 3 2 2 Sample Output Case #1: NO Source 2014 Recommend hujie |
2014多校8 第六题 1006
大意：主角打boss，boss有h血，每回合主角打a血，然后boss回b血。主角打k回合后需要休息1回合，boss也回血。问主角能否打死boss。
题解：
惊天大水题。不过一开始一大堆人wa，我还怕了好久。认真分析，其实是很容易1A的。
我们分3种情况：
1.一刀砍死：
h-a < 1
2.连砍k刀将其砍死：
h-(a-b)*(k-1)-a < 1
3.连砍k刀之后休息一回合，怪的血量减少
h-(a-b)*(k)+b < h
这三种情况就是全部了，其中任意一个为true，就能砍死怪。
题目说可以在没砍到k刀就停下来休息，其实是不优的，要不就一刀砍死，要不就连砍k刀再休息，要是砍不到k刀就休息还能削弱怪物的血，连砍k刀肯定更能削弱怪物的血，这个情况归类到情况3中。
实在是太水了，居然只问yes和no，至少也问一下多少回合砍死啊！
代码：
1 //#pragma comment(linker, "/STACK:102400000,102400000")
2 #include<cstdio>
3 #include<cmath>
4 #include<iostream>
5 #include<cstring>
6 #include<algorithm>
7 #include<cmath>
8 #include<map>
9 #include<set>
10 #include<stack>
11 #include<queue>
12 using namespace std;
13 #define ll long long
14 #define usint unsigned int
15 #define mz(array) memset(array, 0, sizeof(array))
16 #define minf(array) memset(array, 0x3f, sizeof(array))
17 #define REP(i,n) for(i=0;i<(n);i++)
18 #define FOR(i,x,n) for(i=(x);i<=(n);i++)
19 #define RD(x) scanf("%d",&x)
20 #define RD2(x,y) scanf("%d%d",&x,&y)
21 #define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z)
22 #define WN(x) printf("%d\n",x);
23 #define RE freopen("D.in","r",stdin)
24 #define WE freopen("1biao.out","w",stdout)
25
26 int main() {
27 int cas=1, ans;
28 ll h,a,b,k;
29 while(scanf("%I64d%I64d%I64d%I64d",&h,&a,&b,&k)!=EOF) {
30 if(h==0 && a==0 && b==0 && k==0)break;
31 if(h-a<1)ans=1;
32 else if(h-(a-b)*(k-1)-a<1)ans=1;
33 else {
34 if(h-(a-b)*(k)+b>=h)ans=0;
35 else ans=1;
36 }
37 if(ans==0)printf("Case #%d: NO\n",cas++);
38 else printf("Case #%d: YES\n",cas++);
39 }
40 return 0;
41 }
hdu4950 Monster （水题）