pvalue 2 for Parameter colsample_bylevel exceed bound [0,1] colsample_bylevel: Subsample ratio of columns, resample on each level.

时间: 2024-06-06 17:10:58 浏览: 111
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verilog_intro-Cygwin.zip_The Meaning of You_cadence

As an AI language model, I do not have access to the specific context in which this error occurred. However, based on the error message provided, it seems that the parameter "colsample_bylevel" is being assigned a value of 2, which exceeds the allowed range of values (0 to 1). "colsample_bylevel" is a parameter that controls the subsampling of columns at each split level in a tree-based model. To fix this error, the value assigned to "colsample_bylevel" needs to be adjusted to a value between 0 and 1. This can be done by either modifying the code or by passing a valid value as an argument while calling the function.
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Mircea has n pictures. The i-th picture is a square with a side length of si centimeters. He mounted each picture on a square piece of cardboard so that each picture has a border of w centimeters of cardboard on all sides. In total, he used c square centimeters of cardboard. Given the picture sizes and the value c, can you find the value of w? A picture of the first test case. Here c=50=52+42+32, so w=1 is the answer. Please note that the piece of cardboard goes behind each picture, not just the border. Input The first line contains a single integer t (1≤t≤1000) — the number of test cases. The first line of each test case contains two positive integers n (2≤n≤2⋅105) and c (1≤c≤1018) — the number of paintings, and the amount of used square centimeters of cardboard. The second line of each test case contains n space-separated integers si (1≤si≤104) — the sizes of the paintings. The sum of n over all test cases doesn't exceed 2⋅105. Additional constraint on the input: Such an integer w exists for each test case. Please note, that some of the input for some test cases won't fit into 32-bit integer type, so you should use at least 64-bit integer type in your programming language (like long long for C++). Output For each test case, output a single integer — the value of w which was used to use exactly c squared centimeters of cardboard. Example inputCopy 10 3 50 3 2 1 1 100 6 5 500 2 2 2 2 2 2 365 3 4 2 469077255466389 10000 2023 10 635472106413848880 9181 4243 7777 1859 2017 4397 14 9390 2245 7225 7 176345687772781240 9202 9407 9229 6257 7743 5738 7966 14 865563946464579627 3654 5483 1657 7571 1639 9815 122 9468 3079 2666 5498 4540 7861 5384 19 977162053008871403 9169 9520 9209 9013 9300 9843 9933 9454 9960 9167 9964 9701 9251 9404 9462 9277 9661 9164 9161 18 886531871815571953 2609 10 5098 9591 949 8485 6385 4586 1064 5412 6564 8460 2245 6552 5089 8353 3803 3764 outputCopy 1 2 4 5 7654321 126040443 79356352 124321725 113385729 110961227 Note The first test case is explained in the statement. For the second test case, the chosen w was 2, thus the only cardboard covers an area of c=(2⋅2+6)2=102=100 squared centimeters. For the third test case, the chosen w was 4, which obtains the covered area c=(2⋅4+2)2×5=102×5=100×5=500 squared centimeters. c++实现

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这道题目的意思是给定两个长度为n的01字符串a和b，每次操作可以选择一个区间[l, r]，将a中该区间内的元素取反，将b中该区间外的元素取反，问是否能通过有限次操作将a和b都变成全0。 思路：首先考虑每次操作对a和b的...

写出这个题的代码：You are given an unsorted permutation p1,p2,…,pn. To sort the permutation, you choose a constant k (k≥1) and do some operations on the permutation. In one operation, you can choose two integers i, j (1≤j<i≤n) such that i−j=k, then swap pi and pj. What is the maximum value of k that you can choose to sort the given permutation? A permutation is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2,3,1,5,4] is a permutation, but [1,2,2] is not a permutation (2 appears twice in the array) and [1,3,4] is also not a permutation (n=3 but there is 4 in the array). An unsorted permutation p is a permutation such that there is at least one position i that satisfies pi≠i. Input Each test contains multiple test cases. The first line contains the number of test cases t (1≤t≤104). The description of the test cases follows. The first line of each test case contains a single integer n (2≤n≤105) — the length of the permutation p. The second line of each test case contains n distinct integers p1,p2,…,pn (1≤pi≤n) — the permutation p. It is guaranteed that the given numbers form a permutation of length n and the given permutation is unsorted. It is guaranteed that the sum of n over all test cases does not exceed 2⋅105. Output For each test case, output the maximum value of k that you can choose to sort the given permutation. We can show that an answer always exists.

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You are given an array a1,a2,…,an, where all elements are different. You have to perform exactly k operations with it. During each operation, you do exactly one of the following two actions (you choose which to do yourself): find two minimum elements in the array, and delete them; find the maximum element in the array, and delete it. You have to calculate the maximum possible sum of elements in the resulting array. Input The first line contains one integer t (1≤t≤104) — the number of test cases. Each test case consists of two lines: the first line contains two integers n and k (3≤n≤2⋅105; 1≤k≤99999; 2k<n) — the number of elements and operations, respectively. the second line contains n integers a1,a2,…,an (1≤ai≤109; all ai are different) — the elements of the array. Additional constraint on the input: the sum of n does not exceed 2⋅105. Output For each test case, print one integer — the maximum possible sum of elements in the resulting array.

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You are given a positive integer n . In this problem, the MEX of a collection of integers c1,c2,…,ck is defined as the smallest positive integer x which does not occur in the collection c . The primality of an array a1,…,an is defined as the number of pairs (l,r) such that 1≤l≤r≤n and MEX(al,…,ar) is a prime number. Find any permutation of 1,2,…,n with the maximum possible primality among all permutations of 1,2,…,n . Note: A prime number is a number greater than or equal to 2 that is not divisible by any positive integer except 1 and itself. For example, 2,5,13 are prime numbers, but 1 and 6 are not prime numbers. A permutation of 1,2,…,n is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2,3,1,5,4] is a permutation, but [1,2,2] is not a permutation (2 appears twice in the array), and [1,3,4] is also not a permutation (n=3 but there is 4 in the array). Input Each test contains multiple test cases. The first line contains the number of test cases t (1≤t≤104 ). The description of the test cases follows. The only line of each test case contains a single integer n (1≤n≤2⋅105 ). It is guaranteed that the sum of n over all test cases does not exceed 2⋅105 . Output For each test case, output n integers: a permutation of 1,2,…,n that achieves the maximum possible primality. If there are multiple solutions, print any of them. 用c++如何实现，并写出题解

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pythonCalculate A * B. 输入 Each line will contain two integers A and B. Process to end of file. Note: the length of each integer will not exceed 50000.

可以使用 Python 的内置函数 int() 将输入的字符串转换为整数，然后使用乘法运算符 * 计算 A * B。 示例代码如下： while True: try: a, b = map(int, input().split()) print(a * b) except: ...

请给出下题的c++参考代码For an array of integers [a1,a2,…,an], let's call the value |a1−a2|+|a2−a3|+⋯+|an−1−an| the contrast of the array. Note that the contrast of an array of size 1 is equal to 0 . You are given an array of integers a . Your task is to build an array of b in such a way that all the following conditions are met: b is not empty, i.e there is at least one element; b is a subsequence of a, i.e b can be produced by deleting some elements from a (maybe zero); the contrast of b is equal to the contrast of a . What is the minimum possible size of the array b ? Input The first line contains a single integer t (1≤t≤104 ) — the number of test cases. The first line of each test case contains a single integer n (1≤n≤3⋅105) — the size of the array a . The second line contains n integers a1,a2,⋅,an (0≤ai≤109 ) — elements of the array itself. The sum of n over all test cases doesn't exceed 3⋅105 . Output For each test case, print a single integer — the minimum possible size of the array b .

max_diff = max(max_diff, abs(a[i] - a[i - 1]) + abs(a[i + 1] - a[i]) - abs(a[i + 1] - a[i - 1])); } cout << sum - max_diff ; } return 0; } 思路： 题目要求找到一个非空子序列，其 contrast 值...

class Settings(object): DEBUG = False LOG_DIR = None LOG_FILE = "log.txt" RESIZE_METHOD = staticmethod(cocos_min_strategy) # keypoint matching: kaze/brisk/akaze/orb, contrib: sift/surf/brief CVSTRATEGY = ["mstpl", "tpl", "sift", "brisk"] if LooseVersion('3.4.2') < LooseVersion(cv2.__version__) < LooseVersion('4.4.0'): CVSTRATEGY = ["mstpl", "tpl", "brisk"] KEYPOINT_MATCHING_PREDICTION = True THRESHOLD = 0.7 # [0, 1] THRESHOLD_STRICT = None # dedicated parameter for assert_exists OPDELAY = 0.1 FIND_TIMEOUT = 20 FIND_TIMEOUT_TMP = 3 PROJECT_ROOT = os.environ.get("PROJECT_ROOT", "") # for using other script SNAPSHOT_QUALITY = 10 # 1-100 https://pillow.readthedocs.io/en/5.1.x/handbook/image-file-formats.html#jpeg # Image compression size, e.g. 1200, means that the size of the screenshot does not exceed 1200*1200 IMAGE_MAXSIZE = os.environ.get("IMAGE_MAXSIZE", None) SAVE_IMAGE = True

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function [t,s] = specxture(f) %SPECXTURE Computes self-correlation of an image. % [SRAD, SANG, S] = SPECXTURE(F) computes SRAD, the spectral energy % distribution as a function of radius from the center of the % spectrum, SANG, the spectral energy distribution as a function of % angle for 0 to 180 degrees in increments of 1 degree, and S = % log(1 + spectrum of f), normalized to the range [0, 1]. The % maximum value of radius is min(M,N), where M and N are the number % of rows and columns of image (region) f. Thus, SRAD is a row % vector of length = (min(M, N)/2) - 1; and SANG is a row vector of % length 180. s = abs(fft2(f)).^2; %imshow(f,[]);figure;mesh(S); [M, N] = size(s); % Maximum radius that guarantees a circle centered at (x0, y0) that % does not exceed the boundaries of S. rmax = min(M, N); [U,V]=meshgrid(1:rmax,1:rmax); S = s(1:rmax,1:rmax); % Compute isotropy SU = U.^2.*S; SU = sum(SU(:)); SV = V.^2.*S; SV = sum(SV(:)); SUV = U.*V.*S; SUV = sum(SUV(:)); iso = (SU-SV)/sqrt((SU+SV)^2-4.*SUV);%方向 w = 1/sqrt((SU+SV)/sum(S(:)));%相关长度 s = log(s); t(1) = w; t(2) = iso;转为pyhton代码

U, V = np.meshgrid(np.arange(1, rmax + 1), np.arange(1, rmax + 1)) S = s[0:rmax, 0:rmax] SU = np.sum(U ** 2 * S) SV = np.sum(V ** 2 * S) SUV = np.sum(U * V * S) iso = (SU - SV) / np.sqrt((SU + ...

Jonathan is fighting against DIO's Vampire minions. There are n of them with strengths a1,a2,…,an. Denote (l,r) as the group consisting of the vampires with indices from l to r. Jonathan realizes that the strength of any such group is in its weakest link, that is, the bitwise AND. More formally, the strength level of the group (l,r) is defined as f(l,r)=al&al+1&al+2&…&ar. Here, & denotes the bitwise AND operation. Because Jonathan would like to defeat the vampire minions fast, he will divide the vampires into contiguous groups, such that each vampire is in exactly one group, and the sum of strengths of the groups is minimized. Among all ways to divide the vampires, he would like to find the way with the maximum number of groups. Given the strengths of each of the n vampires, find the maximum number of groups among all possible ways to divide the vampires with the smallest sum of strengths. Input The first line contains a single integer t (1≤t≤104) — the number of test cases. The description of test cases follows. The first line of each test case contains a single integer n (1≤n≤2⋅105) — the number of vampires. The second line of each test case contains n integers a1,a2,…,an (0≤ai≤109) — the individual strength of each vampire. The sum of n over all test cases does not exceed 2⋅105. Output For each test case, output a single integer — the maximum number of groups among all possible ways to divide the vampires with the smallest sum of strengths.c++实现

for (int i = 0; i ; i++) { cin >> strengths[i]; } // Sort the strengths in non-decreasing order sort(strengths.begin(), strengths.end()); int minStrength = strengths[0]; int groupCount = 1; ...

Copil Copac is given a list of n−1 edges describing a tree of n vertices. He decides to draw it using the following algorithm: Step 0 : Draws the first vertex (vertex 1 ). Go to step 1 . Step 1 : For every edge in the input, in order: if the edge connects an already drawn vertex u to an undrawn vertex v , he will draw the undrawn vertex v and the edge. After checking every edge, go to step 2 . Step 2 : If all the vertices are drawn, terminate the algorithm. Else, go to step 1 . A reading is defined as the number of times Copil Copac performs step 1 . Find the number of readings needed by Copil Copac to draw the tree. Input Each test contains multiple test cases. The first line of input contains a single integer t (1≤t≤104 ) — the number of test cases. The description of test cases follows. The first line of each test case contains a single integer n (2≤n≤2⋅105 ) — the number of vertices of the tree. The following n−1 lines of each test case contain two integers ui and vi (1≤ui,vi≤n , ui≠vi ) — indicating that (ui,vi) is the i -th edge in the list. It is guaranteed that the given edges form a tree. It is guaranteed that the sum of n over all test cases does not exceed 2⋅105 . Output For each test case, output the number of readings Copil Copac needs to draw the tree.

我们可以将 Copil Copac 进行一次 Step 1 的操作看作 BFS 算法中的一轮搜索。当所有的顶点都被标记为已绘制时，算法终止。 因为每个顶点最多被访问一次，每个边最多被访问两次，所以时间复杂度为 O(n)。每个测试...

For an array of integers [a1,a2,…,an], let's call the value |a1−a2|+|a2−a3|+⋯+|an−1−an| the contrast of the array. Note that the contrast of an array of size 1 is equal to 0 . You are given an array of integers a . Your task is to build an array of b in such a way that all the following conditions are met: b is not empty, i.e there is at least one element; b is a subsequence of a, i.e b can be produced by deleting some elements from a (maybe zero); the contrast of b is equal to the contrast of a . What is the minimum possible size of the array b ? Input The first line contains a single integer t (1≤t≤104 ) — the number of test cases. The first line of each test case contains a single integer n (1≤n≤3⋅105) — the size of the array a . The second line contains n integers a1,a2,⋅,an (0≤ai≤109 ) — elements of the array itself. The sum of n over all test cases doesn't exceed 3⋅105.

$|a_1 - a_2| + |a_2 - a_3| + \cdots + |a_{n-1} - a_n| = |a_1 - a_n| + |a_n - a_{n-1}| + \cdots + |a_2 - a_1|$ 我们可以令 $b$ 为 $a$ 中某个值最大/最小的数，然后在 $a$ 中寻找与 $b$ 相邻的两个数，即 $a_...

You are given two arrays a and b each consisting of n integers. All elements of a are pairwise distinct. Find the number of ways to reorder a such that ai>bi for all 1≤i≤n , modulo 109+7 . Two ways of reordering are considered different if the resulting arrays are different. Input Each test contains multiple test cases. The first line contains the number of test cases t (1≤t≤104 ). The description of the test cases follows. The first line of each test case contains a single integer n (2≤n≤2⋅105 ) — the length of the array a and b . The second line of each test case contains n distinct integers a1 , a2 , … , an (1≤ai≤109 ) — the array a . It is guaranteed that all elements of a are pairwise distinct. The second line of each test case contains n integers b1 , b2 , … , bn (1≤bi≤109 ) — the array b . It is guaranteed that the sum of n over all test cases does not exceed 2⋅105 . Output For each test case, output the number of ways to reorder array a such that ai>bi for all 1≤i≤n , modulo 109+7 .

找出重新排列数组a的方式，使得对于所有1≤i≤n，ai>bi。如果得到的数组不同，则认为两种重新排列方式是不同的。求满足条件的方案数，对109+7取模。 思路：贪心+逆序对 首先，对数组a进行升序排序，对数组b进行...

t = 0:1/800:0.01; window_length = 0.001; num_windows = length(t)/window_length; phi = zeros(1,length(t)); for i = 1:num_windows start_index = round((i-1)window_length800)+1; end_index = round(iwindow_length800); phi(start_index:end_index) = 2pi800t(start_index:end_index); end报错Index exceeds the number of array elements. Index must not exceed 9. 出错 test (第 21 行) phi(start_index:end_index) = 2pi800t(start_index:end_index);

您的代码中，第21行的语句有一个小问题，就是 2pi800t(start_index:end_index) 中的乘号 * 缺失了。应该将其改为 2*pi*800*t(start_index:end_index)。修改后的完整代码如下： t = 0:1/800:0.01; window_...

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