「BZOJ3856」Monster
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
0 0 0 0
Sample Output
Case #1: NO
题解
PoPoQQQ:
题目大意：给定一只喵，初始h点HP，每回合先手砍一刀a点伤害，喵后手回b点血，先手k回合攻击之后休息一次，问先手能否砍死喵
C++语法基础题23333333
很容易WA- –
能砍死的三个充分条件：
1.OTK
2.第k回合砍死
3.休息之后喵的血量比初始低
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 |
#include<iostream> #include<set> #include<map> #include<cstdio> #include<cstring> #include<cstdlib> #include<ctime> #include<vector> #include<queue> #include<algorithm> #include<cmath> #define inf 2000000000 #define pa pair<int,int> #define ll long long using namespace std; ll read() { ll x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } ll h,a,b,K; int cnt; int main() { while(scanf("%lld%lld%lld%lld",&h,&a,&b,&K)) { if(!h&&!a&&!b&&!K)return 0; if(a>=h||K*a-(K-1)*b>=h||K*a>(K+1)*b)printf("Case #%d: YES\n",++cnt); else printf("Case #%d: NO\n",++cnt); } return 0; } |
Subscribe
计数器 9793299
OI 课件=>
OI 课件=>