zoukankan      html  css  js  c++  java
  • HDU-4950 Monster

    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): 151    Accepted Submission(s): 64


    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
    #include<iostream>
    #include<cstdio>
    using namespace std;
    int main()
    {
          int t=1;
         __int64 h,a,b,k;
        while(~scanf("%I64d%I64d%I64d%I64d",&h,&a,&b,&k))
        {
            if(h==0&&a==0&&b==0&&k==0)
                  break;
             printf("Case #%d: ",t++);
    
             if(h<=a)
               {
    
                  printf("YES
    ");
                   continue;
               }
               if(a<=b)
                  {
                      printf("NO
    ");
                      continue;
                  }
            if(h-((a-b)*(k-1)+a)<=0)
               {
    
                  printf("YES
    ");
                   continue;
               }
    
             if(h-(a-b)*k+b>=h)
                {
                    printf("NO
    ");
                    continue;
                }
            else
              {
    
                  printf("YES
    ");
                  continue;
              }
        }
        return 0;
    }
  • 相关阅读:
    14.Java基础_函数/函数重载/参数传递
    98. 验证二叉搜索树(深搜)
    13.Java基础_数组内存图
    12Java基础_数组定义格式/动态初始化/静态初始化
    计算几何基础
    11.Java基础_IDEA常用快捷键
    Add Two Numbers
    Two Sum
    登录界面id属性的使用
    系统查看
  • 原文地址:https://www.cnblogs.com/cancangood/p/3913253.html
Copyright © 2011-2022 走看看