01 模拟 - 竞技编程&基础算法

模拟, 顾名思义, 就是题目让你做什么你就做什么, 它考察的是将思路转换为代码的代码能力, 这类题目一般较为简单, 是比赛里面的签到题

模拟 - 多项式输出

https://www.luogu.com.cn/problem/P1067

这个题讲了一大堆构成多项式的条件, 经过仔细的分析再一一对应的去模拟就行了

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参考题解

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#include <iostream>
using namespace std;
int main()
{
    int n;
    cin >> n;
    bool first = true;
    for (int i = n; i >= 0; i--)
    {
        int m;
        cin >> m;

        if (m == 0) continue;

        if (m < 0)
            cout << "-";
        else if (!first)
            cout << "+";

        first = false;
        m = abs(m);
        if (m != 1 || i == 0)
            cout << m;

        
        if (i == 1)
            cout << "x";
        else if (i != 0)
            cout << "x^" << i;
    }

    return 0;
}

模拟 - 蛇形方阵

https://www.luogu.com.cn/problem/P5731 使用方向向量的方式去解决所有方阵填数字的问题

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参考题解

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#include <iostream>
using namespace std;

const int N = 10;
int arr[N][N];
// 1.方向向量
int pos;
int dx[] = {0, 1, 0, -1};
int dy[] = {1, 0, -1, 0};

int main()
{
    int n;
    cin >> n;
    int cnt = 1;
    int x = 1, y = 1;
    while (cnt <= n * n)
    {
        arr[x][y] = cnt;

        //  2.判断边界
        int nx = x + dx[pos];
        int ny = y + dy[pos];
        if (nx < 1 || nx > n || ny < 1 || ny > n || arr[nx][ny])
        {
            pos = (pos + 1) % 4;
            nx = x + dx[pos];
            ny = y + dy[pos];
        }
        x = nx;
        y = ny;
        cnt++;
    }

    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= n; j++)
        {
            printf("%3d", arr[i][j]);
        }
        cout << "\n";
    }
    return 0;
}

模拟 - 字符串展开

https://www.luogu.com.cn/problem/P1098 纯模拟 - 考察代码能力

参考题解

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#include <iostream>
#include <string>
#include <algorithm>
using namespace std;
int p1, p2, p3;
string s;

bool sameType(char a, char b)
{
    return (islower(a) && islower(b)) || (isdigit(a) && isdigit(b));
}

void changeDash(string::iterator i)
{
    string dash;
    for (char k = (*(i - 1) + 1); k < *(i + 1); k++)
    {
        for (int j = 1; j <= p2; j++)
            dash += k;
    }

    for (auto &e : dash)
    {
        if (p1 == 1)
            e = tolower(e);
        else if (p1 == 2)
            e = toupper(e);
        else
            e = '*';
    }

    if (p3 == 2)
        reverse(dash.begin(), dash.end());

    cout << dash;
}

int main()
{
    cin >> p1 >> p2 >> p3 >> s;
    for (auto i = s.begin(); i != s.end(); ++i)
    {
        if (*i == '-' 
            && i != s.begin() 
            && i + 1 != s.end() 
            && *(i - 1) < *(i + 1) 
            && sameType(*(i - 1), *(i + 1)))
            changeDash(i);
        else
            cout << *i;
    }
    return 0;
}