Codeforces Round #890 (Div. 2)

A.Tales of a Sort

题目大意

Alphen has an array of positive integers $a$ of length n.

Alphen can perform the following operation:

• For all $i$ from 1 to n, replace $a_i$ with $\max (0,a_i-1)$ .

Alphen will perform the above operation until $a$ is sorted, that is $a$ satisfies $a_1\le a_2\le \dots \le a_n$. How many operations will Alphen perform? Under the constraints of the problem, it can be proven that Alphen will perform a finite number of operations.

思路

记录最大的逆序对的值即可

#include <bits/stdc++.h>
using namespace std;
const int N = 550;
int a[N];

void solve()
{
   
    int n;
    cin >> n;
    for (int i = 1; i <= n; i++)
        cin >> a[i];
    int ans = 0;
    for (int i = 1; i <= n; i++)
        for (int j = i + 1; j <= n; j++)
        {
   
            if (a[i] > a[j])
                ans = max(ans, a[i]);
        }
    cout << ans << endl;
}
int main()
{
   
    int t;
    cin >> t;
    while (t--)
        solve();
    return 0;
}

B.Good Arrays

题目大意

Let’s call an array of positive integers b of length n good if:

1. $a_i\ne b_i$ for all i from $1ton$ ,
2. $a_1+a_2+\dots +a_n=b_1+b_2+\dots +b_n$.

思路

这道题将非 $1$ 的数把值累加到 $1$ 上面，让自己等于 $1$, 然后 $1$ 变成非 $1$ 。

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
void solve()
{
   
    int n;
    cin >> n;
    LL cnt1 = 0, sum = 0;

    for (int i = 1; i <= n; i++)
    {
   
        int x;
        scanf("%d", &x);
        if (x == 1)
            cnt1++;
        else
            sum += (x - 1);
    }
    if (n == 1)
    {
   
        puts("NO");
        return;
    }
    if (sum >= cnt1)
    {
   
        puts("YES");
        return;
    }
    else
    {
   
        puts("NO");
        return;
    }
}

int main()
{
   
    int t;
    cin >> t;
    while (t--)
        solve();
    return 0;
}