v4t 1507(题意 水题)

Execution of Paladin
Description

Murloc is a powerful race in Hearthstone. In the set League of Explorers, a new Paladin ability card called Anyfin Can Happen is released with the ability to summon 7 Murlocs that died this game. If there aren’t enough dead Murlocs, it may summon less than 7 Murlocs.
There are many different minions in Murloc race, here are four of them:

Coldlight Oracle: 3 MANA, 2 ATTACK, 2 HP. Battlecry: Each player draws 2 cards.

Murloc Warleader: 3 MANA, 3 ATTACK, 3 HP. ALL other Murlocs have +2/+1.

Bluegill Warrior: 2 MANA, 2 ATTACK, 3 HP. Charge.

Old Murk-Eye: 4 MANA, 2 ATTACK, 3 HP. Charge. Has +1 Attack for each other Murloc on the Battlefield.
Here are some explanations:

MANA: The cost of summon the minion. Minions summoned by ability cards cost no mana besides the cost of the ability cards. Every player has 10 MANAs at most.

ATTACK: How many damage can the minion make once.

HP: How many attacks can the minion or heroes take.

Battlecry: An ability where a particular effect activates when the card with the Battlecry is played directly from the hand. The minions summoned by ability won’t activate their Battlecry.

Charge: Minions cannot attack at once when they are summoned unless they have Charge description. They will have to wait until next turn.

Battlefield: The battlefield (or game board) is where the action takes place, representing the board on which each game is played out.

+2/+1: +2 ATTACK and +1 HP.

Now, it is your turn. You have 10 MANAs and only one card: Anyfin Can Happen. There are nothing on the Battlefield, which means your minions can directly attack enemy hero. You can remember the list of dead Murlocs. You know how many HP the enemy hero remains. Will you win this game through this only card you have?

Input

							 Multiple test cases. The first line contains an integer T (T<= 22000), indicating the number of test case.

The first line of each test contains two integers, n (the number of dead Murlocs, 0 <= n <= 7) and h (the HP of enemy hero, 0 < h <= 30).

Then n lines follows, each line contains a string, indicates the name of dead Murloc. The string will only be “Coldlight Oracle”, “Murloc Warleader”, “Bluegill Warrior” or “Old Murk-Eye”.

Output

One line per case. If you can win the game in this turn, output “Mrghllghghllghg!”(Without quotes). Otherwise, output “Tell you a joke, the execution of Paladin.” You will win the game if you attack enemy hero with your minions and make his/her HP less or equal than 0.

Sample

Input

Copy
3
3 1
Coldlight Oracle
Coldlight Oracle
Murloc Warleader
3 8
Old Murk-Eye
Old Murk-Eye
Coldlight Oracle
7 30
Old Murk-Eye
Bluegill Warrior
Bluegill Warrior
Murloc Warleader
Murloc Warleader
Coldlight Oracle
Coldlight Oracle

Output
CopyTell you a joke, the execution of Paladin.
Mrghllghghllghg!
Tell you a joke, the execution of Paladin.

Hint

In the first test case, none of the Murlocs can attack.

In the second test case, every Old Murk-Eye has +2 ATTACK because there is another Old Murk-Eye and a Coldlight Oracle. So the total damage is 8.

In the last test case, Old Murk-Eye has 12 ATTACK (2 basic ATTACK, 6 other Murlocs and 2 Murloc Warleader), two Bluegill Warriors has 6 ATTACK(2 basic ATTACK, and 2 Murloc Warleader) each. So the total damage is 24.

#include<bits/stdc++.h>
using namespace std;
int main()
{
   
    int t,n,ans;
    char tmp[15000];
    scanf("%d",&t);
    while(t--)
    {
   
        int a[4]= {
   3,2,2,2};
        int b[4]= {
   0,0,0,0};
        scanf("%d %d",&n,&ans);
        getchar();
        for(int i=0; i<n; i++)
        {
   
            gets(tmp);
            if(tmp[0]=='O')
            {
   
                b[2]+=1;
            }
            if(tmp[0]=='B')
            {
   
                b[1]+=1;
            }
            if(tmp[0]=='M')
            {
   
                b[0]+=1;
            }
        }
        a[1]=a[1]*b[1]+b[1]*b[0]*2;
        a[2]=a[2]*b[2]+b[2]*b[0]*2+b[2]*(n-1);
        if(ans>a[1]+a[2])
        {
   
            cout<<"Tell you a joke, the execution of Paladin."<<endl;
        }
        else
        {
   
            cout<<"Mrghllghghllghg!"<<endl;
        }
    }
    return 0;
}

全部评论

相关推荐

会飞的猿:我看你想进大厂,我给你总结一下学习路线吧,java语言方面常规八股要熟,那些java的集合,重点背hashmap八股吧,jvm类加载机制,运行时分区,垃圾回收算法,垃圾回收器CMS、G1这些,各种乐观锁悲观锁,线程安全,threadlocal这些。在进阶一些的比如jvm参数,内存溢出泄漏排查,jvm调优。我这里说的只是冰山一角,详细八股可以去网上找,这不用去买,都免费资源。mysql、redis可以去看小林coding,我看你简历上写了,你一定要熟,什么底层b+树、索引结构、innodb、mvcc、undo log、redo log、行级锁表级锁,这些东西高频出现,如果面试官问我这些我都能笑出来。消息队列rabbitmq也好kafka也好,学一种就行,什么分区啊副本啊确认机制啊怎么保证不重复消费、怎么保证消息不丢失这些基本的一定要会,进阶一点的比如LEO、高水位线、kafka和rocketmq底层零拷贝的区别等等。计算机网络和操作系统既然你是科班应该理解起来问题不大,去看小林coding这两块吧,深度够了。spring boot的八股好好看看吧,一般字节腾讯不这么问,其他的java大厂挺爱问的,什么循环依赖啥的去网上看看。数据结构的话科班应该问题不大,多去力扣集中突击刷题吧。项目的话其实说白了还是结合八股来,想一想你写的这些技术会给你挖什么坑。除此之外,还有场景题、rpc、设计模式、linux命令、ddd等。不会的就别往简历上写了,虽然技术栈很多的话好看些,但背起来确实累。总结一下,多去实习吧,多跳槽,直到跳到一个不错的中厂做跳板,这是一条可行的进大厂的路线。另外,只想找个小厂的工作的话,没必要全都照这些准备,太累了,重点放在框架的使用和一些基础八股吧。大致路线就这样,没啥太多难度,就是量大,你能达到什么高度取决于你对自己多狠,祝好。
点赞 评论 收藏
分享
评论
点赞
收藏
分享

创作者周榜

更多
牛客网
牛客企业服务