【Acwing】The 62nd Weekly Game Solution

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AcWing 4500. 三个元素

输入样例1：

6
3 1 4 1 5 9

输出样例1：

4 1 3

输入样例2：

5
1 1000000000 1 1000000000 1

输出样例2：

-1 -1 -1

输入样例3：

9
10 10 11 10 10 10 10 10 1

输出样例3：

9 8 3

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方法一：Read in and find first max 和 min 值 ,Then loop again to find any ,The final judgment can be made

#include <iostream>
#include <algorithm>

using namespace std;

const int N = 3010, INF = 0x3f3f3f3f;

int a[N];

int main()
{
    int max = 0, min = INF, mid = 0;
    int i1  = -1, i2  = -1 ,  i3  = -1;
    
    int n; cin >> n;
    
    for(int i = 1; i <= n; i ++ )
    {
        cin >> a[i];
        if(a[i] > max) max = a[i], i1 = i;
        if(a[i] < min) min = a[i], i2 = i;
    }
    
    for(int i = 1; i <= n; i ++ )
    {
        if(a[i] > min && a[i] < max) mid = a[i], i3 = i;
    }
    
    //printf("%d %d %d\n", min, max, mid);
    
    if(max != min && max != mid && min != mid &&
       i1 != i2 && i1 != i3 && i2 != i3 &&
       i1 != -1 && i2 != -1 && i3 != -1)
    {
        printf("%d %d %d", i2, i3, i1);
    }
    else puts("-1 -1 -1");
    
    return 0;
}

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方法2：使用mapto store and judge 

#include <iostream>
#include <cstring>
#include <map>
#include <vector>

using namespace std;

int main()
{
    int n; cin >> n;
    map<int, int> pos;
    
    for(int i = 1; i <= n; i ++ )
    {
        int x; cin >> x;
        pos[x] = i;
    }
    
    if(pos.size() < 3) puts("-1 -1 -1");
    else 
    {
        vector<int> res;
        for(auto [k, v] : pos) res.push_back(v);
        for(int i = 0; i < 3; i ++ )
            cout << res[i] << ' ';
    }
    
    return 0;
}

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 AcWing 4501. 收集卡牌

输入样例1：

3 11
2 3 1 2 2 2 3 2 2 3 1

输出样例1：

00100000001

输入样例2：

4 8
4 1 3 3 2 3 3 3

输出样例2：

00001000

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

using namespace std;

const int N = 100010;

int n, m;
int cnt[N];

int main()
{
    cin >> n >> m;
    int tot = 0;
    
    while(m -- )
    {
        int x; scanf("%d", &x);
        if(!cnt[x]) tot ++ ;
        cnt[x] ++ ;
        
        if(tot == n)
        {
            printf("1");
            for(int i = 1; i <= n; i ++ )
                if(-- cnt[i] == 0)
                    tot -- ;
        }
        else
        {
            printf("0");
        }
    }
    
    return 0;
}

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AcWing 4502. 集合操作 

输入样例1：

6
1 3
2
1 4
2
1 8
2

输出样例1：

0.000000
0.500000
3.000000

输入样例2：

4
1 1
1 4
1 5
2

输出样例2：

2.000000

---

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

using namespace std;

const int N = 500010;

int n, m;
int w[N];

int main()
{
    scanf("%d", &m);

    double res = 0, sum = 0;
    int k = 0;
    while (m -- )
    {
        int op, x;
        scanf("%d", &op);
        if (op == 1)
        {
            scanf("%d", &x);
            w[ ++ n] = x;
            while (k + 1 <= n && w[k + 1] <= (sum + w[n]) / (k + 1))
                sum += w[ ++ k];
            res = max(res, w[n] - (sum + w[n]) / (k + 1));
        }
        else
        {
            printf("%lf\n", res);
        }
    }

    return 0;
}