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++;elsesum += (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;
}

C.To Become Max

题目描述

You are given an array of integers $a$ of length $n$ .

In one operation you:

• Choose an index $i$ such that $1\le i\le n-1$ and $a_i\le a_{i+1}$ .
• Increase $a_i$ by $1$ .

Find the maximum possible value of $\max (a_1,a_2,\dots a_n)$ that you can get after performing this operation at most $k$ times.

输入格式

Each test contains multiple test cases. The first line of input contains a single integer $t$ ( $1\le t\le 100$ ) — the number of test cases. The description of the test cases follows.

The first line of each test case contains two integers $n$ and $k$ ( $2\le n\le 1000$ , $1\le k\le 10^8$ ) — the length of the array $a$ and the maximum number of operations that can be performed.

The second line of each test case contains $n$ integers $a_1,a_2,\dots ,a_n$ ( $1\le a_i\le 10^8$ ) — the elements of the array $a$ .

It is guaranteed that the sum of $n$ over all test cases does not exceed $1000$ .

输出格式

For each test case output a single integer — the maximum possible maximum of the array after performing at most $ k $ operations.

样例 #1

样例输入 #1

6
3 4
1 3 3
5 6
1 3 4 5 1
4 13
1 1 3 179
5 3
4 3 2 2 2
5 6
6 5 4 1 5
2 17
3 5

样例输出 #1

4
7
179
5
7
6

提示

In the first test case, one possible optimal sequence of operations is: $[1,3,3]\to [2,3,3]\to [2,4,3]\to [3,4,3]\to [4,4,3]$ .

In the second test case, one possible optimal sequence of operations is:
$[1,3,4,5,1]\to [1,4,4,5,1]\to [1,5,4,5,1]\to [1,5,5,5,1]\to [1,5,6,5,1]\to [1,6,6,5,1]\to [1,7,6,5,1]$

思路

我们二分答案，下界设为 $0$，而 $\max (a_1,\dots a_n)+k$ 作为上界。

设 $b$ 是 $a$ 在执行 $k$ 次操作后的数组。假设对于一个数 $x$，我们想要检查是否可以在 $k$ 次操作中得到$\max (b_1,\dots b_n)\ge x$。也就是说，通过 $k$ 次操作使得存在某个索引 $i$ 使得 $b_i\ge x$ 。

因此，让我们从遍历一下 $b$ 数组 ，看看是否可以在 $k$ 次操作中使得 $b_i\ge x$ 。

设 $f(i,y)$ 为使得 $b_i\ge y$ 所需的最小操作数。那么有：
$$\begin{gathered} f(i,y)=0,对于所有y\le ai \\ f(i,y)=y-a_i+f(i+1,y-1)，对于所有1\le a_i \\ f(i,y)=+\infty ，对于i=n且所有y>a_i。 \end{gathered}$$
显然计算 $f(i,x)$ 最坏情况下需要 $O(n)$ 的时间。因此，我们看看 ${\sum }_{i=1}^n{f}_{i,x}\le k$

如果操作次数 $\le k$，那么在 $k$ 次操作中可能存在某个 $b_i\ge x$，并在更新当前答案后让 $l=mid$。否则这个 $x$ 没有办法满足，让 $r=mid$。复杂度：$O(n^2\log A)$，其中 $A$ 是 $a_i$ 和 $k$ 的最大可能值。

我们这里再来看

一个优化空间的思路，因为我们只需要计算所有 $f$ 值的和，所以 $\text{O(1)}$ 空间复杂度就可以了。

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 2005;
vector<ll> a(N);
ll n, k;
bool check(ll x)
{for (int i = 1; i <= n; i++){vector<ll> minneed(n + 1);minneed[i] = x;ll ut = 0;for (int j = i; j <= n; j++){if (minneed[j] <= a[j])break;if (j == n){ut = k + 1;break;}ut += minneed[j] - a[j];minneed[j + 1] = max(0ll, minneed[j] - 1);}if (ut <= k)return true;}return false;
}void solve()
{cin >> n >> k;ll maxnum = 0;for (int i = 1; i <= n; i++){cin >> a[i];maxnum = max(maxnum, a[i]);}// cout << *max_element(a.begin(), a.end()) + k + 1 << endl;ll l = -1, r = maxnum + k + 1, ans = 0;// cout << r << endl;while (l + 1 < r){ll mid = (l + r) >> 1;if (check(mid))ans = mid, l = mid;elser = mid;}cout << l << endl;
}int main()
{int t;cin >> t;while (t--)solve();return 0;
}