A. Everything Everywhere All But One

思路：找数组的平均值是否在数组的元素中

代码如下：

#include <bits/stdc++.h>

using namespace std;

typedef pair<int, int> PII;
typedef pair<char, int> PCI;
typedef long long LL;

const int N = 1010;

int T;

void solve()
{
    int n, m;
    scanf("%d", &n);

    int a[N];
    int sum = 0;
    for(int i = 0; i < n; i ++ )
        scanf("%d", &a[i]), sum += a[i];

    for(int i = 0; i < n; i ++ )
    {
        if( fabs ( sum*1.0/n - 1.0*a[i]) < 1e-6)
        {
            puts("YES");
            return;
        }
    }

    puts("NO");

    return;
}

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

    while(T -- )
        solve();

    return 0;
}

B. Odd Subarrays

思路：

找到一个逆序对，res ++，输出即可

代码如下：

#include <bits/stdc++.h>

using namespace std;

typedef pair<int, int> PII;
typedef pair<char, int> PCI;
typedef long long LL;

const int N = 2e5+10;

int T;

void solve()
{
    int n, m;
    scanf("%d", &n);

    int a[N];
    int sum = 0;
    for(int i = 0; i < n; i ++ )
        scanf("%d", &a[i]);

    int res = 0;
    for(int i = 1; i < n; i ++ )
    {
        if(a[i-1] > a[i]) res ++, i ++;
    }

    cout << res << endl;

    return;
}

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

    while(T -- )
        solve();

    return 0;
}

C. Circular Local MiniMax

思路：将后半段插入到前半段中，n 必须是偶数，若是奇数，插入的后半段或前半段必定会有两个同一区域的数挨着，我们必须满足两端区域的数是交替出现如此可满足

代码如下：

#include <bits/stdc++.h>

using namespace std;

typedef pair<int, int> PII;
typedef pair<char, int> PCI;
typedef long long LL;

const int N = 2e5+10;

int T;

void solve()
{
    int n, m;
    scanf("%d", &n);

    int a[N];
    for(int i = 1; i <= n; i ++ )
        scanf("%d", &a[i]);

    sort(a + 1, a + n + 1);

    int res[N], cnt = 1;

    for(int i = 1, j = n/2 + 1; i <= n/2 || j <= n; i ++ , j ++ )
    {
        if(i <= n/2) res[cnt ++] = a[i];
        if(j <= n) res[cnt++] = a[j];
    }

    res[0] = res[n];
    res[n+1] = res[1];

    for(int i = 1; i <= n; i ++ )
    {
        if((res[i-1] >= res[i] && res[i+1] <= res[i]) || (res[i-1] <= res[i] && res[i+1] >= res[i]))
        {
            //printf("%d %d %d
", res[i-1], res[i], res[i+1]);
            puts("NO");
            return;
        }
    }

    puts("YES");
    for(int i = 1; i <= n; i ++ )
    {
        printf("%d ", res[i]);
    }

    puts("");

    return;
}

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

    while(T -- )
        solve();

    return 0;
}