打卡 思路：哈希查找

#include<iostream>
using namespace std;
#define IOS ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define N 1010
int w[N] = {0};

int main()
{
    IOS;
    int n, m;
    cin >> n >> m;
    for (int i = 0;i < n;i++)
    {
        cin >> w[0];
        if (w[0] < 1000) w[w[0]]++;
    }
    for (int i = 1;i < m / 2;i++)
    {
        w[i]--;
        w[m - i]--;
        if (w[i] >=0 && w[m - i] >= 0)
        {
            cout << i << " " << m - i << endl;
            return 0;
        }
    }
    cout << "No Solution" << endl;
    return 0;
}

打卡，二分法 浮点

#include<iostream>
using namespace std;
#define IOS ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define N 100010

int n, m;
int w[N];

bool check(double mid)
{
    int cnt = 0;
    for (int i = 0;i < n;i++)
        cnt += w[i] / mid;
    return cnt >= m;
}

int main()
{
    IOS;
    cin >> n >> m;
    for (int i = 0;i < n;i++) cin >> w[i];

    double l = 0, r = 1e9;
    for (int i = 0;i < 100;i++)
    {
        double mid = (l + r) / 2;
        if (check(mid)) l = mid;
        else r = mid;
    }
    printf("%.2lf\n", r);
    return 0;
}

复杂度O(n) 打卡打卡

#include<iostream>
#include<string>
using namespace std;
#define IOS ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);

int main()
{
    IOS;
    string s1, s2;
    cin >> s1 >> s2;
    int num = s1.length(), sum = 0;
    for (int i = 0;i < num - 1;i++)
    {
        if (s1[i] != s2[i])
        {
            s1[i] = s2[i];
            if (s1[i + 1] == '*') s1[i + 1] = 'o';
            else s1[i + 1] = '*';
            sum++;
        }
    }
    cout << sum;
    return 0;
}

打卡打卡

#include<iostream>
#include<algorithm>
using namespace std;
#define IOS ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
typedef long long ll;

struct Student
{
    int Score;
    int Chinese;
    int Num;
}Stu[305];

int cmp(Student s1, Student s2)
{
    if (s1.Score != s2.Score)
        return s1.Score > s2.Score;
    else if (s1.Chinese != s2.Chinese)
        return s1.Chinese > s2.Chinese;
    else
        return s1.Num < s2.Num;
}

int main()
{
    IOS;
    int n;
    cin >> n;
    int C, M, E;
    for (int i = 1;i <= n;i++)
    {
        cin >> C >> M >> E;
        Stu[i].Score = C + M + E;
        Stu[i].Chinese = C;
        Stu[i].Num = i;
    }
    sort(Stu + 1, Stu + 1 + n, cmp);
    for (int i = 1;i <= 5;i++)
        cout << Stu[i].Num << " " << Stu[i].Score << endl;
    return 0;
}

打卡

#include<bits/stdc++.h>
using namespace std;
#define IOS ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
typedef long long ll;
int s[10005] = {};

int main()
{
    IOS;
    int l, m;
    int begin ,end;
    cin >> l >> m;
    int size = sizeof(s[0]);
    for (int i = 0;i < m;i++)
    {
        cin >> begin >> end;
        memset(&s[begin], -1, (end - begin +1) * size);
    }
    int sum = 0;
    for (int i = 0;i <= l; i++)
    {
        if (s[i] == 0)sum++;
    }
    cout << sum;
    return 0;
}

每日一题第五天打卡

#include <bits/stdc++.h>
using namespace std;

//该函数参考了博客https://www.cnblogs.com/zwjjj/p/9953718.html
string intToA(int n,int radix)    //n是待转数字，radix是指定的进制
{
    string ans="";
    do{
        int t=n%radix;
        if(t>=0&&t<=9)    ans+=t+'0';
        else ans+=t-10+'A';
        n/=radix;
    }while(n!=0);    //使用do{}while（）以防止输入为0的情况
    reverse(ans.begin(),ans.end());
    return ans;    
}

int main()
{
    int b;
    string s;
    cin >> b;
    for ( int i=1;i<=300;i++ )
    {
        s=intToA(i*i,b);
        int n=s.length();
        int j;
        for( j=0;j<n/2 && s[j]==s[n-j-1];j++ );
        if( j==n/2 )
        {
            cout << intToA(i,b) << " " << s <<endl;
        }
    }
    return 0;
}

每日一题第五天打卡

#include <bits/stdc++.h>
using namespace std;

//该函数参考了博客https://www.cnblogs.com/zwjjj/p/9953718.html
string intToA(int n,int radix)    //n是待转数字，radix是指定的进制
{
    string ans="";
    do{
        int t=n%radix;
        if(t>=0&&t<=9)    ans+=t+'0';
        else ans+=t-10+'A';
        n/=radix;
    }while(n!=0);    //使用do{}while（）以防止输入为0的情况
    reverse(ans.begin(),ans.end());
    return ans;    
}

int main()
{
    int b;
    string s;
    cin >> b;
    for ( int i=1;i<=300;i++ )
    {
        s=intToA(i*i,b);
        int n=s.length();
        int j;
        for( j=0;j<n/2 && s[j]==s[n-j-1];j++ );
        if( j==n/2 )
        {
            cout << intToA(i,b) << " " << s <<endl;
        }
    }
    return 0;
}

每日一题第四天打卡，这题不会，参考猿六的代码，学习了一下

#include<bits/stdc++.h>
using namespace std;

const int N = 25;

int n, m;
char g[N][N];//存储地板
bool st[N][N];//存储状态，该位置地板是否被走过

int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};//上右下左四个方向

int dfs(int x, int y)//深度优先遍历
{
    int cnt = 1;

    st[x][y] = true;
    for (int i = 0; i < 4; i ++ )//走四个方向
    {
        int a = x + dx[i], b = y + dy[i];
        if (a < 0 || a >= n || b < 0 || b >= m) continue;
        if (g[a][b] != '.') continue;
        if (st[a][b]) continue;

        cnt += dfs(a, b);//如果可以走向某一个方向，对该方向上的点递归
    }

    return cnt;
}

int main()
{
    while (cin >> m >> n, n || m)
    {
        for (int i = 0; i < n; i ++ ) cin >> g[i];//一次读入一行

        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;
                }

        memset(st, 0, sizeof st);//清空状态数组
        cout << dfs(x, y) << endl;
    }

    return 0;
}
//参考了猿六