Hdu 4950 Monster

Hdu 4950 Monster

Monster Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission (s): 590 Accepted Submission (s): 238


Problem DescriptionTeacher 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 blocked by. 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 ".
InputThere 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 ).
OutputFor 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 20 0 0 0

Sample Output

Case #1: NO

Source2014 Multi-University Training Contest 8

Question and code:

Sign-in questions, not much to talk about, divided into three stages of simulation, the first stage, to see if it can be a moment to kill the monster; the second stage, to the k round, suppose I conduct an attack, the monster has not returned blood to see if it can be killed;

In the third stage, play the k + 1 round to see if the monsters have lost blood in these rounds.

#include 
 
  #include 
  
   using namespace std;int main(){    int cas=1;    long long h,m,n,t;    while(scanf("%I64d%I64d%I64d%I64d",&h,&m,&n,&t)!=EOF)    {        if(!h&&!m&&!n&&!t)            break;        if(m<=n)        {            if(h<=m)            {                printf("Case #%d: YES\n",cas++);            }            else            {                printf("Case #%d: NO\n",cas++);            }            continue;        }        if(m*t-n*(t-1)>=h)        {            printf("Case #%d: YES\n",cas++);        }        else        {           if(m*t-n*(t+1)>=1)            printf("Case #%d: YES\n",cas++);           else            printf("Case #%d: NO\n",cas++);        }    }    return 0;}