HDOJ 4950 Monster

最新推荐文章于 2019-12-22 09:17:22 发布

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最新推荐文章于 2019-12-22 09:17:22 发布

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分类专栏： ACM 那些年我们一起做过的ACM 文章标签： 2014 Multi-Universit

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本文探讨了教师玛伊如何通过合理的行动策略，在特定条件下击败HP受到初始值、攻击减益和增益影响的怪物。通过解析输入参数并应用逻辑判断，实现对怪物HP的有效控制，最终决定能否成功击杀怪物。

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题目：

Monster

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 198 Accepted Submission(s): 81

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

Sample Output

  
  
   
   Case #1: NO

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解题思路：

如果a>=h，直接可以打死怪兽，否则，只有在第k次打死怪兽或者是没k+1次都可以对怪兽造成伤害，才能打死怪兽。

代码：

#include <cstdio>
#include <cstring>

typedef long long lint;

int  main()
{
    int h, a, b, k, icase = 1;
    while(~scanf("%d%d%d%d", &h, &a, &b, &k))
    {
        if(0==h && 0==a && 0== b && 0==k)
            break;
        bool flag = false;
        if(a >= h) flag = true;
        else
        {
            lint t = 1ll*h - 1ll*(k-1)*(a-b) - 1ll*a;
            if(t <= 0) flag = true;
            t = 1ll*a*k - 1ll*b*(k+1);
            if(t > 0) flag = true;
        }
        printf("Case #%d: %s\n", icase++, flag ? "YES" : "NO");
    }
    return 0;
}