寒假第一周：每日一题打卡

货仓选址

中位数法

#include <iostream>
#include <math.h>
#include <algorithm>
using namespace std;
const int N=1e5+5;
int main()
{
    int n;
    cin>>n;
    int a[N];
    for(int i=0;i<n;i++)
    {
        cin>>a[i];
    }
    sort(a,a+n);
    int res=0;
    for(int i=0;i<n;i++)
        res+=abs(a[i]-a[n>>1]);//这里改成i>>1也可以AC
    cout<<res<<endl;
    return 0;
}

红与黑

洪水灌溉、

//宽搜写法
#include <iostream>
#include <queue>
#include <algorithm>

#define x first
#define y second

using namespace std;

typedef pair<int, int> PII;
const int N = 25;

int n, m;
char g[N][N];

int bfs(int sx, int sy)
{
    queue<PII> q;
    q.push({sx, sy});
    g[sx][sy] = '#';
    int res = 0;

    int dx[] = {-1, 0, 1, 0}, dy[] = {0, 1, 0, -1};

    while (q.size())
    {
        auto t = q.front();
        q.pop();
        res ++ ;

        for (int i = 0; i < 4; i ++ )
        {
            int x = t.x + dx[i], y = t.y + dy[i];
            if (x < 0 || x >= n || y < 0 || y >= m || g[x][y] != '.') continue;
            g[x][y] = '#';
            q.push({x, y});
        }
    }
    return res;
}

int main()
{
    while (cin >> m >> n, n || m)
    {
        for (int i = 0; i < n; i ++ ) cin >> g[i];
        int x, y;
        for (int i = 0; i < n; i ++ )
            for (int j = 0; j < m; j ++ )
                if (g[i][j] == '@')
                {
                    x = i;
                    y = j;
                }
        cout << bfs(x, y) << endl;
    }

    return 0;
}

//深搜写法
#include <iostream>

using namespace std;

const int N = 25;

int n, m;
char g[N][N];
int dx[] = {-1, 0, 1, 0}, dy[] = {0, 1, 0, -1};

int dfs(int x, int y)
{
    int res = 1;
    g[x][y] = '#';
    for (int i = 0; i < 4; i ++ )
    {
        int a = x + dx[i], b = y + dy[i];
        if (a >= 0 && a < n && b >= 0 && b < m && g[a][b] == '.')
            res += dfs(a, b);
    }
    return res;
}

int main()
{
    while (cin >> m >> n, n || m)
    {
        for (int i = 0; i < n; i ++ ) cin >> g[i];
        int x, y;
        for (int i = 0; i < n; i ++ )
            for (int j = 0; j < m; j ++ )
                if (g[i][j] == '@')
                {
                    x = i;
                    y = j;
                }
        cout << dfs(x, y) << endl;
    }

    return 0;
}

回文平方

考点：
1.十进制转化成其他进制：短除法
2.其他进制转化成十进制：秦九韶算法
3.其他进制转化成其他进制
(1) 用十进制过渡
(2) 短除法
4.双指针法判断是否是回文数

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
char get(int x)
{
    if(x <= 9) return x +'0';
    return x-10+'A';
}
string change(int n,int b)
{
    string num;
    while(n) num+=get(n % b),n/=b;
    reverse(num.begin(),num.end());
    return num;
}
bool check(string num)
{
    for(int i=0,j=num.size() - 1;i<j;i++,j--)
    {
        if(num[i]!=num[j])return false;
    }
    return true;
}
int main()
{
    int b;
    cin>>b;
    for(int i=1;i<=300;i++)
    {
        auto ans=change(i*i,b);
        if(check(ans))
        {
            cout<<change(i,b)<<' '<<ans<<endl;
        }
    }
    return 0;
}

寒假第一周每日一题打卡，希望这个寒假可以不负众望