202009-2 风险人群筛查

• C++
• 总结

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本题链接：202009-2 风险人群筛查

本博客给出本题截图：

C++

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

int n, k, t, x1, y1, x2, y2;

int main()
{
    cin >> n >> k >> t >> x1 >> y1 >> x2 >> y2;

    int res1 = 0, res2 = 0;
    while (n -- )
    {
        bool r1 = false, r2 = false;
        int s = 0;
        for (int i = 0; i < t; i ++ )
        {
            int x, y;
            cin >> x >> y;
            if (x >= x1 && x <= x2 && y >= y1 && y <= y2)
            {
                s ++ ;
                r1 = true;
                if (s >= k) r2 = true;
            }
            else s = 0;
        }
        if (r1) res1 ++ ;
        if (r2) res2 ++ ;
    }
    cout << res1 << endl << res2 << endl;
    return 0;
}

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总结

水题，不解释