1116 Come on! Let's C (20 分)(hash散列)

"Let's C" is a popular and fun programming contest hosted by the College of Computer Science and Technology, Zhejiang University. Since the idea of the contest is for fun, the award rules are funny as the following:

  1. The Champion will receive a "Mystery Award" (such as a BIG collection of students' research papers...).
  2. Those who ranked as a prime number will receive the best award -- the Minions (小黄人)!
  3. Everyone else will receive chocolates.

Given the final ranklist and a sequence of contestant ID's, you are supposed to tell the corresponding awards.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (<=10000), the total number of contestants. Then N lines of the ranklist follow, each in order gives a contestant's ID (a 4-digit number). After the ranklist, there is a positive integer K followed by K query ID's.

Output Specification:

For each query, print in a line "ID: award" where the award is "Mystery Award", or "Minion", or "Chocolate". If the ID is not in the ranklist, print "Are you kidding?" instead. If the ID has been checked before, print "ID: Checked".

Sample Input:

6
1111
6666
8888
1234
5555
0001
6
8888
0001
1111
2222
8888
2222

Sample Output:

8888: Minion
0001: Chocolate
1111: Mystery Award
2222: Are you kidding?
8888: Checked
2222: Are you kidding?

生词

英文 解释
College 学院
department
Champion 冠军
Mystery 神秘事物
Minions 小黄人
ranklist 用户排名
contestant 比赛者
query 查询

分析:

ran数组标记每个id对应的排名,集合ss存储所有已经询问过的id,如果发现当前id已经出现在ss中,则输出“Checked”,如果ran[id] == 0说明当前id不在排名列表中,所以输出“Are you kidding?”,如果ran[id]为1则输出“Minion”,如果ran[id]为素数则输出“Mystery Award”,否则输出“Chocolate”

原文链接:https://blog.csdn.net/liuchuo/article/details/52505009

题解

自己用哈希做的,感觉也比较简单,对于不同的情况标记不同的值即可。

#include <bits/stdc++.h>

using namespace std;
bool prime(int n)
{
    if(n<=1) return false;
    for(int i=2;i*i<=n;i++){
        if(n%i==0) return false;
    }
    return true;
}
int rk[10010];
int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("1.txt", "r", stdin);
#endif
    int n,k,data;
    cin>>n;
    for(int i=0; i<n; i++)
    {
        cin>>data;
        rk[data]=i+1;
    }
    cin>>k;
    for(int i=0; i<k; i++)
    {
        cin>>data;
        if(rk[data]==1){
            printf("%04d",data);
            cout<<": Mystery Award"<<endl;
            rk[data]=-1;
        }else if(prime(rk[data])){
            printf("%04d",data);
            cout<<": Minion"<<endl;
            rk[data]=-1;
        }else if(rk[data]==-1){
            printf("%04d",data);
            cout<<": Checked"<<endl;
        }else if(rk[data]==0){
            printf("%04d",data);
            cout<<": Are you kidding?"<<endl;
        }else{
            printf("%04d",data);
            cout<<": Chocolate"<<endl;
            rk[data]=-1;
        }
    }
    return 0;
}
posted @ 2022-01-07 17:33  勇往直前的力量  阅读(31)  评论(0编辑  收藏  举报