Codeforces Round 887 (Div. 2) 题解||A+B

A. Desorting

Call an array $a$ of length $n$ sorted if $a_1\le a_2\le \dots \le a_{n-1}\le a_n$.

Ntarsis has an array $a$ of length $n$.

He is allowed to perform one type of operation on it (zero or more times):

• Choose an index $i$ ($1\le i\le n-1$).
• Add $1$ to $a_1,a_2,\dots ,a_i$.
• Subtract $1$ from $a_{i+1},a_{i+2},\dots ,a_n$.

The values of $a$ can be negative after an operation.

Determine the minimum operations needed to make $a$ not sorted.
Input

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

The first line of each test case contains a single integer $n$ ($2\le n\le 500$) — the length of the array $a$.

The next line contains $n$ integers $a_1,a_2,\dots ,a_n$ ($1\le a_i\le 10^9$) — the values of array $a$.

It is guaranteed that the sum of $n$ across all test cases does not exceed $500$.
Output

Output the minimum number of operations needed to make the array not sorted.
Example
input

4
2
1 1
4
1 8 10 13
3
1 3 2
3
1 9 14

output

1
2
0
3

Note

In the first case, we can perform $1$ operation to make the array not sorted:

• Pick $i=1$. The array $a$ then becomes $[2,0]$, which is not sorted.

In the second case, we can perform $2$ operations to make the array not sorted:

• Pick $i=3$. The array $a$ then becomes $[2,9,11,12]$.
• Pick $i=3$. The array $a$ then becomes $[3,10,12,11]$, which is not sorted.

It can be proven that $1$ and $2$ operations are the minimal numbers of operations in the first and second test cases, respectively.

In the third case, the array is already not sorted, so we perform $0$ operations.

题解

1.题意

首先是题目大致的意思是给你一个长度为N的整数序列，他去使用一个操作去使整个序列不是排序的，整个操作可以概括为选取一个下标为i 的让前 i 项的值都加上 1 后面的项都减去一。

2.思路

寻找一段两点之间的区间，让他们的值保证是最小的，（简化），对这个值进行操作。

3.代码

#include <iostream>
#include <cmath>
using namespace std ;
const int N = 510 ;
int q[N] ;
int ans ;
int main ()
{

	int t ;
	cin >> t ;
	while ( t -- )
	{
		ans = 1e9;//赋值方式： 1， 1en + m 
	int n ;
	cin >> n ;
	for(int i = 0 ; i < n ; i ++ )
	{
		cin >> q[i] ;
		if(i != 0 )
		{
			ans  = min(ans , q[i] - q[i - 1]);
		}
	}
	if(ans < 0 )
		cout << 0 << endl ; 
	else cout << ( ans / 2 )+ 1 << endl ; 
}	
	return 0 ; 
}
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B. Fibonaccharsis

Ntarsis has received two integers $n$ and $k$ for his birthday. He wonders how many fibonacci-like sequences of length $k$ can be formed with $n$ as the $k$-th element of the sequence.

A sequence of non-decreasing non-negative integers is considered fibonacci-like if $f_i=f_{i-1}+f_{i-2}$ for all $i>2$, where $f_i$ denotes the $i$-th element in the sequence. Note that $f_1$ and $f_2$ can be arbitrary.

For example, sequences such as $[4,5,9,14]$ and $[0,1,1]$ are considered fibonacci-like sequences, while $[0,0,0,1,1]$, $[1,2,1,3]$, and $[-1,-1,-2]$ are not: the first two do not always satisfy $f_i=f_{i-1}+f_{i-2}$, the latter does not satisfy that the elements are non-negative.

Impress Ntarsis by helping him with this task.

Input

The first line contains an integer $t$ ($1\le t\le 2\cdot 10^5$), the number of test cases. The description of each test case is as follows.

Each test case contains two integers, $n$ and $k$ ($1\le n\le 2\cdot 10^5$, $3\le k\le 10^9$).

It is guaranteed the sum of $n$ over all test cases does not exceed $2\cdot 10^5$.

Output

For each test case output an integer, the number of fibonacci-like sequences of length $k$ such that the $k$-th element in the sequence is $n$. That is, output the number of sequences $f$ of length $k$ so $f$ is a fibonacci-like sequence and $f_k=n$. It can be shown this number is finite.

Example
input

8
22 4
3 9
55 11
42069 6
69420 4
69 1434
1 3
1 4

output

4
0
1
1052
11571
0
1
0

Note

There are $4$ valid fibonacci-like sequences for $n=22$, $k=4$:

• $[6,8,14,22]$,
• $[4,9,13,22]$,
• $[2,10,12,22]$,
• $[0,11,11,22]$.

For $n=3$, $k=9$, it can be shown that there are no fibonacci-like sequences satisfying the given conditions.

For $n=55$, $k=11$, $[0,1,1,2,3,5,8,13,21,34,55]$ is the only fibonacci-like sequence.

题解

1，思路

在作者自身的题解中使用了二分搜素，但是这个题暴力也能过，斐波那契数列，他给定我们一个 n 也就是最后一项，这个时候我们要反向的思考数学性质，毕竟这是一道B 题不可能太难。

因为斐波那契数列的S₁ +S₂ =S₃ 这个性质，根据递归的思想我们只需要再确定一项就能确定完整的数列，也就是一层递归，我们再用一层k的遍历来验证答案的正确性即可（数学简化）

3.代码

#include <iostream>
using namespace std ;
bool cnt ;
int main () 
{
	int t ;
	cin >> t ;
	while (t --)
	{
		int n , k ;
		int ans = 0 ;

		cin >> n >> k ; 
		for(int i = n / 2 ; i <= n ; i ++)
		{	
			cnt = false ;
			int ans1 = n ;
			int ans2 = i ; 
			int kk = k - 2 ;
			for(int j = n - i ;   kk > 0; j = ans1 - ans2 ,kk -- )
			{
				ans1 = ans2 ;
				ans2 = j ;
				if(ans1 < ans2 || ans1 < 0 || ans2 < 0)
				{
					cnt = true;
					break;
				}
			}
			if(!cnt)
				ans ++ ;
		}
		cout << ans << endl ;
	}
	return 0 ;
}
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