PAT 甲级 1116. Come on! Let's C (20) 【循环判断】

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本文提供了一道 PAT A 类竞赛题 1116 的详细解答过程，通过 C++ 实现了一个 ID 查询系统。该系统能够根据不同条件返回 MysteryAward、Minion、Chocolate 或 Areyoukidding? 等反馈。代码中运用了 map 容器进行数据存储及查询。

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题目链接

https://www.patest.cn/contests/pat-a-practise/1116

思路
注意一个细节
如果没有本来 ID 的后来又查了这个ID 不是输出 checked 而是 Are you kidding?

AC代码

#include <cstdio>
#include <cstring>
#include <ctype.h>
#include <cstdlib>
#include <cmath>
#include <climits>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <map>
#include <stack>
#include <set>
#include <numeric>
#include <sstream>
#include <iomanip>
#include <limits>

#define CLR(a) memset(a, 0, sizeof(a))
#define pb push_back

using namespace std;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
typedef pair <int, int> pii;
typedef pair <ll, ll> pll;

const double PI  = 3.14159265358979323846264338327;
const double E   = exp(1);
const double eps = 1e-6;

const int INF  = 0x3f3f3f3f;
const int maxn = 1e5 + 5;
const int MOD  = 1e9 + 7;

bool isprime(int x)
{
    int m = sqrt(x) + 1;
    for (int i = 2; i <= m; i++)
    {
        if (x % i == 0)
            return false;
    }
    return true;
}

int main()
{
    int n, num;
    map <int, int> m, flag;
    cin >> n;
    for (int i = 1; i <= n; i++)
    {
        scanf("%d", &num);
        m[num] = i;
    }
    cin >> n;
    for (int i = 0; i < n; i++)
    {
        scanf("%d", &num);
        printf("%04d: ", num);
        int vis = m[num];
        if (vis == 0)
            printf("Are you kidding?\n");
        else if (flag[num] == 1)
            printf("Checked\n");
        else if (vis == 1)
            printf("Mystery Award\n");
        else if (isprime(vis))
            printf("Minion\n");
        else if (m[num])
            printf("Chocolate\n");
        flag[num] = 1;
    }
}