hdu 4950 Monster--2014 Multi-University Training Contest 8

最新推荐文章于 2024-04-19 08:20:14 发布

原创最新推荐文章于 2024-04-19 08:20:14 发布 · 886 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#2014 Multi-Universit #acm #算法 #签到题

算法同时被 2 个专栏收录

115 篇文章

订阅专栏

C++

79 篇文章

订阅专栏

本文介绍了一个关于Teacher Mai如何通过攻击减少怪物HP直至其死亡的问题。分析了不同参数下Teacher Mai能否成功击败怪物的情况，并提供了相应的代码实现。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

链接：http://acm.hdu.edu.cn/showproblem.php?pid=4950

Monster

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 260 Accepted Submission(s): 114

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

Recommend

hujie | We have carefully selected several similar problems for you: 4955 4954 4953 4951 4949

Statistic | Submit | Discuss | Note

这是这场的签到题。没啥好解释的。不过情况稍微有点复杂，一个个分析就完了。

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
int main(){
    int kase=0;
    long long h,a,b,k;
    while(cin>>h>>a>>b>>k){
        if(h==0 && a==0 && b==0 && k==0){
            break;
        }
        if(h<=a){
            printf("Case #%d: YES\n",++kase);
            continue;
        }
        if(a<=b){
            printf("Case #%d: NO\n",++kase);
            continue;
        }
        if((a-b)*k>=(h-b)){
            printf("Case #%d: YES\n",++kase);
            continue;
        }
        if((a-b)*k-b>0){
            printf("Case #%d: YES\n",++kase);
            continue;
        }
        else{
            printf("Case #%d: NO\n",++kase);
        }
    }
    return 0;
}