Monster

Time 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 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

Source

2014 Multi-University Training Contest 8

题解及代码：

签到题，不多讲，分成三个阶段模拟，第一阶段，看是否能一下就把怪物杀死；第二阶段，进行到第k回合，假设我进行攻击，怪物未回血，看是否能将其杀死；

第三阶段，进行玩第k+1回合，看这几个回合下来怪物是否掉血了。

#include <iostream>
#include <cstdio>
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;
}

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