hdu4950 Monster

Monster

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1782 Accepted Submission(s): 550

Problem 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
  
  

Author

xudyh

题意：已知你的攻击力a、怪物血量h、每回合怪物的回血量b和每k回合你必须休息一回合；你也可以不到k回合就休息。

分析：只需要判断每k回合怪物掉的血量比回的血量多就YES，否则就NO。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <cmath>
#include <algorithm>
using namespace std;
const double eps = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
#define ll long long
#define CL(a) memset(a,0,sizeof(a))

int main()  
{  
    ll h,a,b,k,cas=1;  
    while(~scanf("%I64d%I64d%I64d%I64d",&h,&a,&b,&k),h+a+b+k)  
    {  
        if(a>=h||(a-b)*k-b>0||(a-b)*(k-1)+a>=h)//只要判断每个k回合怪物掉的血比回的血多久行
            printf("Case #%lld: YES\n",cas++);  
        else  
            printf("Case #%lld: NO\n",cas++);  
    }  
    return 0;  
}