CF1692G 2^Sort 题解

题目

链接

https://www.luogu.com.cn/problem/CF1692G

字面描述

题面翻译

给你一个长度为 $n(\sum n<2\cdot 10^5)$ 的数组 $a$，问你在这个数组中，有多少个长度为 $k+1(1\le k<n)$ 的区间，符合以下的条件：

$$\begin{gathered} 2^0\cdot a_i<2^1\cdot a_{i+1}<2^2\cdot a_{i+2}<\cdots <2^k\cdot a_{i+k} \\ 注：i为这个区间开始的位置 \end{gathered}$$

由tzyt翻译

题目描述

Given an array $ a $ of length $ n $ and an integer $ k $ , find the number of indices $ 1 \leq i \leq n - k $ such that the subarray $ [a_i, \dots, a_{i+k}] $ with length $ k+1 $ (not with length $ k $ ) has the following property:

• If you multiply the first element by $ 2^0 $ , the second element by $ 2^1 $ , …, and the ( $ k+1 $ )-st element by $ 2^k $ , then this subarray is sorted in strictly increasing order.

More formally, count the number of indices $ 1 \leq i \leq n - k $ such that $ $KaTeX parse error: Can't use function '$' in math mode at position 84: …\cdot a_{i+k}. $̲ $

输入格式

The first line contains an integer $ t $ ( $ 1 \leq t \leq 1000 $ ) — the number of test cases.

The first line of each test case contains two integers $ n $ , $ k $ ( $ 3 \leq n \leq 2 \cdot 10^5 $ , $ 1 \leq k < n $ ) — the length of the array and the number of inequalities.

The second line of each test case contains $ n $ integers $ a_1, a_2, \dots, a_n $ ( $ 1 \leq a_i \leq 10^9 $ ) — the elements of the array.

The sum of $ n $ across all test cases does not exceed $ 2 \cdot 10^5 $ .

输出格式

For each test case, output a single integer — the number of indices satisfying the condition in the statement.

样例 #1

样例输入 #1

6
4 2
20 22 19 84
5 1
9 5 3 2 1
5 2
9 5 3 2 1
7 2
22 12 16 4 3 22 12
7 3
22 12 16 4 3 22 12
9 3
3 9 12 3 9 12 3 9 12

样例输出 #1

2
3
2
3
1
0

提示

In the first test case, both subarrays satisfy the condition:

• $ i=1 $ : the subarray $ [a_1,a_2,a_3] = [20,22,19] $ , and $ 1 \cdot 20 < 2 \cdot 22 < 4 \cdot 19 $ .
• $ i=2 $ : the subarray $ [a_2,a_3,a_4] = [22,19,84] $ , and $ 1 \cdot 22 < 2 \cdot 19 < 4 \cdot 84 $ .

In the second test case, three subarrays satisfy the condition: - $ i=1 $ : the subarray $ [a_1,a_2] = [9,5] $ , and $ 1 \cdot 9 < 2 \cdot 5 $ .

• $ i=2 $ : the subarray $ [a_2,a_3] = [5,3] $ , and $ 1 \cdot 5 < 2 \cdot 3 $ .
• $ i=3 $ : the subarray $ [a_3,a_4] = [3,2] $ , and $ 1 \cdot 3 < 2 \cdot 2 $ .
• $ i=4 $ : the subarray $ [a_4,a_5] = [2,1] $ , but $ 1 \cdot 2 = 2 \cdot 1 $ , so this subarray doesn’t satisfy the condition.

思路

对原数组进行模拟下标$x(2n)$
根据题目
若$a_{x-1}<2\cdot a_x$
$\therefore f_x=1$
否则,$f_x=0$

对$f数组$进行前缀和预处理，就可以实现$\mathrm{O}(n)$时间复杂度的统计了。

代码实现

#include<bits/stdc++.h>
using namespace std;const int maxn=2e5+10;
int t,n,k,ans;
int a[maxn],f[maxn];
int main(){scanf("%d",&t);while(t--){scanf("%d%d",&n,&k);int op=0;ans=0;memset(a,0,sizeof(a));for(int i=1;i<=n;i++){int x;scanf("%d",&x);if(2*x>op)a[i]=1;op=x;}for(int i=1;i<=n;i++){f[i]=f[i-1]+a[i];if(i-k-1<0)continue;if(f[i]-f[i-k]==k)++ans;}printf("%d\n",ans);}return 0;
}