### 源码：git clone https://github.com/qqwweee/keras-yolo3.git

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keras yolov3 tiny_yolo_body网络结构改为vgg16结构

Treasure Map 是如何实现的
Problem Description "Take 147 steps due north, turn 63 degrees clockwise, take 82 steps, ...". Most people don't realize how important accuracy is when following the directions on a pirate's treasure map. If you're even a tiny bit off at the start, you'll end up far away from the correct location at the end. Pirates therefore use very exact definitions. One step, for instance, has been defined by the 1670 Pirate Convention to be exactly two times the size of the wooden leg of Long John Silver, or 1.183 m in metricunits. Captain Borbassa was thus not at all worried when he set sail to the treasure island, having a rope with knots in it, exactly one step apart, for accurately measuring distances. Of course he also brought his good old geotriangle, once given to him by his father when he was six years old. However, on closer inspection of the map, he got an unpleasant surprise. The map was made by the famous captain Jack Magpie, who was notorious for including little gems into his directions.In this case, there were distances listed such as sqrt(33) steps. How do you measure that accurately? Fortunately, his first mate Pythagor came to the rescue. After puzzling for a few hours, he came up with the following solution: let pirate A go 4 steps into the perpendicular direction, and hold one end of the measuring rope there. Then pirate B goes into the desired direction while letting the rope slide through his fingers, until he is exactly 7 steps away from pirate A. Pythagor worked out a formula that states that pirate B has then traveled exactly sqrt(33) steps. Captain Borbassa was impressed, but he revealed that there were more such distances on the map. Paranoid as he is, he refuses to let Pythagor see the map, or even tell him what other distances there are on it. They are all square roots of integers, that's all he gets to know. Only on the island itself will the captain reveal the numbers, and then he expects Pyhtagor to quickly work out the smallest two integer numbers of steps that can combine to create the desired distance, using the method described above. Pythagor knows this is not easy, so he has asked your help. Can you help him by writing a program that can determine these two integers quickly? By the way, he did ask the captain how large the numbers inside the square root could get, and the captain replied "one billion". He was probably exaggerating, but you'd better make sure the program works. If you can successfully help the pirates, you'll get a share of the treasure. It might be gold, it might be silver, or it might even be... a treasure map! Input The first line of the input contains a single number: the number of test cases to follow. Each test case has the following format: 1.One line with one integer N, satisfying 1 <= N <= 10^9. Output For every test case in the input, the output should contain two nonnegative integers, separated by a space, on a single line: the distance pirate A needs to head in the perpendicular direction, and the final distance between pirate A and B, such that pirate B has traveled sqrt(N) steps. If there are multiple solutions, give the one with the smallest numbers. If there are no solutions, the output should be "IMPOSSIBLE" (without the quotation marks) on a single line. Sample Input 4 33 16 50 101 Sample Output 4 7 0 4 IMPOSSIBLE 50 51
tiny yolo 训练 已放弃 (核心已转储)
jerrylew@jerrylew-CW15:~/darknet\$ ./darknet detector train ./cfg/voc.data cfg/tiny-yolo-voc.cfg tiny-yolo-voc layer filters size input output 0 conv 16 3 x 3 / 1 416 x 416 x 3 -> 416 x 416 x 16 1 max 2 x 2 / 2 416 x 416 x 16 -> 208 x 208 x 16 2 conv 32 3 x 3 / 1 208 x 208 x 16 -> 208 x 208 x 32 3 max 2 x 2 / 2 208 x 208 x 32 -> 104 x 104 x 32 4 conv 64 3 x 3 / 1 104 x 104 x 32 -> 104 x 104 x 64 5 max 2 x 2 / 2 104 x 104 x 64 -> 52 x 52 x 64 6 conv 128 3 x 3 / 1 52 x 52 x 64 -> 52 x 52 x 128 7 max 2 x 2 / 2 52 x 52 x 128 -> 26 x 26 x 128 8 conv 256 3 x 3 / 1 26 x 26 x 128 -> 26 x 26 x 256 9 max 2 x 2 / 2 26 x 26 x 256 -> 13 x 13 x 256 10 conv 512 3 x 3 / 1 13 x 13 x 256 -> 13 x 13 x 512 11 max 2 x 2 / 1 13 x 13 x 512 -> 13 x 13 x 512 12 conv 1024 3 x 3 / 1 13 x 13 x 512 -> 13 x 13 x1024 13 conv 1024 3 x 3 / 1 13 x 13 x1024 -> 13 x 13 x1024 14 conv 305 1 x 1 / 1 13 x 13 x1024 -> 13 x 13 x 305 15 detection darknet: ./src/parser.c:281: parse_region: Assertion `l.outputs == params.inputs' failed. 已放弃 (核心已转储) 这什么情况啊 求解
linux内核编译时出错 不知道怎么弄
scripts/kconfig/conf -s arch/mips/Kconfig CHK include/linux/version.h CHK include/linux/utsrelease.h SYMLINK include/asm -> include/asm-mips CALL scripts/checksyscalls.sh CHK include/linux/compile.h /opt/linux-2.6.32/scripts/gen_initramfs_list.sh: Cannot open '/mnt/ramdisk.tiny4/' /opt/linux-2.6.32/usr/Makefile:64: recipe for target 'usr/initramfs_data.cpio.gz' failed make[1]: *** [usr/initramfs_data.cpio.gz] Error 1 Makefile:878: recipe for target 'usr' failed make: *** [usr] Error 2
Ink Blots 怎么来做的
Description Drops of dark ink can fall on a white piece of paper creating a number of round ink blots. Three examples are shown above. The blots can create multiple distinct white regions. In the first figure, there is just one white region. In the second figure there is the outer white region plus a small white region bounded by the left four blots and an even smaller white region bounded by the right three blots. In the third figure, there are four white regions, one on the very outside, one inside the outer ring of blots and outside the four blots in the middle, and two tiny ones each formed between three of the four inner blots. Two points are in the same white region if a path can be drawn between them that only passes through white points. Your problem is to count the number of white regions given the centers and radii of the blots. Math Formulas: If circles C1 with center (x1, y1) and radius r1, and C2 with center (x2, y2) and radius r2 intersect in exactly two distinct points, let d equal the distance between the centers of C1 and C2, A = atan2(y2-y1, x2-x1), and B = acos((r12 + d2 - r22)/(2*r1*d)); then the intersection points on C1 are at angles A+B and A-B radians counterclockwise from the ray extending to the right from the center of C1. The function atan2 is the inverse tangent function with two arguments, and acos is the inverse cosine function, both available in the math libraries of C, C++, and Java. Input There are from one to 15 data sets, followed by a final line containing only 0. A data set starts with a line containing a single positive integer n, which is no more than 100. It is the number of blots in the dataset. Then 3n positive integers follow, with a single blank or a newline separating them. Each group of three give the data for the circular boundary of one blot: x and y coordinate of the center of the blot and its radius, in that order. Each of these numbers will be no larger than 1,000,000. All blots lie entirely on a piece of paper, and no blot touches any edge of the paper. No two circles in a dataset will be identical. Given any two distinct circles, they will either intersect at exactly two distinct points or not intersect at all. If two circles in the input intersect, then they overlap by at least one unit. More precisely, if they have radii r1 and r2, where r1 <= r2, and if d is the distance between their centers, then r2 - r1 + 1 <= d <= r1 + r2 - 1. Three or more circles will never intersect at the same point. If C is a circle in the input that intersects at least one other input circle, and p and q are any of the intersection points of C with any of the other input circles, with p distinct from q, then p and q will be separated on C by at least 0.001 radians of arc. The restrictions on radii and angles ensure that standard double-precision arithmetic is sufficient for the calculations suggested above. The sample input below corresponds to the figures above, though the scale is different in each figure. Output The output contains one line for each data set. The line contains only the number of white regions for the dataset, which is never more than 200. Sample Input 4 45 45 40 65 55 35 45 45 10 20 95 10 5 30 30 20 30 60 20 60 30 20 60 60 20 90 45 15 16 200 120 65 300 100 55 400 120 65 480 200 65 500 300 55 480 400 65 400 480 65 300 500 55 200 480 65 120 400 65 100 300 55 120 200 65 300 245 60 300 355 60 385 300 51 215 300 51 0 Sample Output 1 3 4
Shoring Up the Levees 线段坐标的计算问题
ATMEL Tiny13 单片机检验字节问题
ATMEL Tiny13这款单片机的校验字节是否能修改，还有，校验字节有什么用？ 这两个字节是否会导致同一程序下载到不同的mcu中时数据发送变化？

C的程序语言，Showstopper
Problem Description Data-mining huge data sets can be a painful and long lasting process if we are not aware of tiny patterns existing within those data sets. One reputable company has recently discovered a tiny bug in their hardware video processing solution and they are trying to create software workaround. To achieve maximum performance they use their chips in pairs and all data objects in memory should have even number of references. Under certain circumstances this rule became violated and exactly one data object is referred by odd number of references. They are ready to launch product and this is the only showstopper they have. They need YOU to help them resolve this critical issue in most efficient way. Can you help them? Input Input file consists from multiple data sets separated by one or more empty lines. Each data set represents a sequence of 32-bit (positive) integers (references) which are stored in compressed way. Each line of input set consists from three single space separated 32-bit (positive) integers X Y Z and they represent following sequence of references: X, X+Z, X+2*Z, X+3*Z, …, X+K*Z, …(while (X+K*Z)<=Y). Your task is to data-mine input data and for each set determine weather data were corrupted, which reference is occurring odd number of times, and count that reference. Output For each input data set you should print to standard output new line of text with either “no corruption” (low case) or two integers separated by single space (first one is reference that occurs odd number of times and second one is count of that reference). Sample Input 1 10 1 2 10 1 1 10 1 1 10 1 1 10 1 4 4 1 1 5 1 6 10 1 Sample Output 1 1 no corruption 4 3
bugzilla安装后注册新用户报错
Software error: Can't locate object method "message" via package "SMTP auth requires MIME::Base64 and Authen::SASL at C:/Perl/site/lib/Email/Sender/Transport/SMTP.pm line 162. Email::Sender::Transport::SMTP::_smtp_client(Email::Sender::Transport::SMTP::Persistent=HASH(0x48559bc)) called at C:/Perl/site/lib/Email/Sender/Transport/SMTP/Persistent.pm line 32 Email::Sender::Transport::SMTP::Persistent::_smtp_client(Email::Sender::Transport::SMTP::Persistent=HASH(0x48559bc)) called at C:/Perl/site/lib/Email/Sender/Transport/SMTP.pm line 202 Email::Sender::Transport::SMTP::send_email(Email::Sender::Transport::SMTP::Persistent=HASH(0x48559bc), Email::Abstract=ARRAY(0x484623c), HASH(0x4853064)) called at C:/Perl/site/lib/Email/Sender/Role/CommonSending.pm line 45 Email::Sender::Role::CommonSending::try {...} () called at C:/Perl/lib/Try/Tiny.pm line 103 eval {...} called at C:/Perl/lib/Try/Tiny.pm line 94 Try::Tiny::try(CODE(0x485d714), Try::Tiny::Catch=REF(0x3deeba4)) called at C:/Perl/site/lib/Email/Sender/Role/CommonSending.pm line 58 Email::Sender::Role::CommonSending::send(Email::Sender::Transport::SMTP::Persistent=HASH(0x48559bc), Email::Abstract=ARRAY(0x484623c), HASH(0x485d504)) called at C:/Perl/site/lib/Email/Sender/Simple.pm line 119 Email::Sender::Simple::send_email("Email::Sender::Simple", Email::Abstract=ARRAY(0x484623c), HASH(0x4845f24)) called at C:/Perl/site/lib/Email/Sender/Role/CommonSending.pm line 45 Email::Sender::Role::CommonSending::try {...} () called at C:/Perl/lib/Try/Tiny.pm line 103 eval {...} called at C:/Perl/lib/Try/Tiny.pm line 94 Try::Tiny::try(CODE(0x3c1eb14), Try::Tiny::Catch=REF(0x3c3634c)) called at C:/Perl/site/lib/Email/Sender/Role/CommonSending.pm line 58 Email::Sender::Role::CommonSending::send("Email::Sender::Simple", Bugzilla::MIME=HASH(0x48539c4), HASH(0x3f4452c)) called at C:/Perl/lib/Sub/Exporter/Util.pm line 18 Sub::Exporter::Util::__ANON__(Bugzilla::MIME=HASH(0x48539c4), HASH(0x3f4452c)) called at Bugzilla/Mailer.pm line 212 eval {...} called at Bugzilla/Mailer.pm line 212 Bugzilla::Mailer::MessageToMTA("From: shaviation_fae\@163.com\x{a}To: 13817976834\@163.com\x{a}Subject:"..., 1) called at Bugzilla/Token.pm line 129 Bugzilla::Token::issue_new_user_account_token("jyuan", "13817976834\@163.com") called at Bugzilla/User.pm line 2498 Bugzilla::User::check_and_send_account_creation_confirmation(Bugzilla::User=HASH(0xa3b63c), "jyuan", "13817976834\@163.com") called at C:/bugzilla-5.1.1/createaccount.cgi line 46 " (perhaps you forgot to load "SMTP auth requires MIME::Base64 and Authen::SASL at C:/Perl/site/lib/Email/Sender/Transport/SMTP.pm line 162. Email::Sender::Transport::SMTP::_smtp_client(Email::Sender::Transport::SMTP::Persistent=HASH(0x48559bc)) called at C:/Perl/site/lib/Email/Sender/Transport/SMTP/Persistent.pm line 32 Email::Sender::Transport::SMTP::Persistent::_smtp_client(Email::Sender::Transport::SMTP::Persistent=HASH(0x48559bc)) called at C:/Perl/site/lib/Email/Sender/Transport/SMTP.pm line 202 Email::Sender::Transport::SMTP::send_email(Email::Sender::Transport::SMTP::Persistent=HASH(0x48559bc), Email::Abstract=ARRAY(0x484623c), HASH(0x4853064)) called at C:/Perl/site/lib/Email/Sender/Role/CommonSending.pm line 45 Email::Sender::Role::CommonSending::try {...} () called at C:/Perl/lib/Try/Tiny.pm line 103 eval {...} called at C:/Perl/lib/Try/Tiny.pm line 94 Try::Tiny::try(CODE(0x485d714), Try::Tiny::Catch=REF(0x3deeba4)) called at C:/Perl/site/lib/Email/Sender/Role/CommonSending.pm line 58 Email::Sender::Role::CommonSending::send(Email::Sender::Transport::SMTP::Persistent=HASH(0x48559bc), Email::Abstract=ARRAY(0x484623c), HASH(0x485d504)) called at C:/Perl/site/lib/Email/Sender/Simple.pm line 119 Email::Sender::Simple::send_email("Email::Sender::Simple", Email::Abstract=ARRAY(0x484623c), HASH(0x4845f24)) called at C:/Perl/site/lib/Email/Sender/Role/CommonSending.pm line 45 Email::Sender::Role::CommonSending::try {...} () called at C:/Perl/lib/Try/Tiny.pm line 103 eval {...} called at C:/Perl/lib/Try/Tiny.pm line 94 Try::Tiny::try(CODE(0x3c1eb14), Try::Tiny::Catch=REF(0x3c3634c)) called at C:/Perl/site/lib/Email/Sender/Role/CommonSending.pm line 58 Email::Sender::Role::CommonSending::send("Email::Sender::Simple", Bugzilla::MIME=HASH(0x48539c4), HASH(0x3f4452c)) called at C:/Perl/lib/Sub/Exporter/Util.pm line 18 Sub::Exporter::Util::__ANON__(Bugzilla::MIME=HASH(0x48539c4), HASH(0x3f4452c)) called at Bugzilla/Mailer.pm line 212 eval {...} called at Bugzilla/Mailer.pm line 212 Bugzilla::Mailer::MessageToMTA("From: shaviation_fae\@163.com\x{a}To: 13817976834\@163.com\x{a}Subject:"..., 1) called at Bugzilla/Token.pm line 129 Bugzilla::Token::issue_new_user_account_token("jyuan", "13817976834\@163.com") called at Bugzilla/User.pm line 2498 Bugzilla::User::check_and_send_account_creation_confirmation(Bugzilla::User=HASH(0xa3b63c), "jyuan", "13817976834\@163.com") called at C:/bugzilla-5.1.1/createaccount.cgi line 46 "?) at Bugzilla/Mailer.pm line 214. For help, please send mail to the webmaster (admin@aeroer.com), giving this error message and the time and date of the error.

Problem Description There is an equilateral triangle consist of 3 mirrors. There is a tiny slit in the corners of the triangle, which can let a laser beam pass through. We label the 3 slits as A, B and C. There only exists two ways (see the picture above) make a laser beam enter our triangle though C, reflects 11 times in ther triangle, and exit from the triangle though C. The 2 ways are symmetry. Here is the question for you, our great programmer. How many ways we can make a laser beam enter the triangle though C and exit though C, and the beam reflects n times in the triangle? e.g. there are 80840 ways when n is 1000001. Input The first line contains a number T.(1≤T≤100) Then the following T line each line contains a number n.(1≤n≤107) Output For each n, print the corresponding result. Sample Input 10 2 5 7 11 13 17 19 23 29 31 Sample Output 0 0 2 2 2 0 4 4 2 6
Android下RTL8192cu WiFi模块的移植

using namespace std; const double TINY_VALUE = 1e-10; //计算精度为 10^-10 double tsin(double x) { double g = 0; double t = x; int n = 1; do { g += t; n++; t = -t * x*x / (2 * n - 1) / (2 * n - 2); } while (fabs(t) >= TINY_VALUE); return g; } int main() { double k, r, s; cout << "r = "; cin >> r; cout << "s = "; cin >> s; if (r*r <= s * s) k = sqrt(tsin(r)*tsin(r) + tsin(s)*tsin(s)); else k = tsin(r*s) / 2; cout << k << endl; return 0; } 请问do~while语句里面的代码是什么意思。小白求教，谢谢了。

./darknet detect ./cfg/tiny-yolo-voc.cfg tiny-yolo-voc.weights ./data/eagle.jpg 在我输入这条指令测试时冒出了 layer filters size input output 0 conv 32 3 x 3 / 1 416 x 416 x 3 -> 416 x 416 x 32 1 max 2 x 2 / 2 416 x 416 x 32 -> 208 x 208 x 32 2 conv 64 3 x 3 / 1 208 x 208 x 32 -> 208 x 208 x 64 3 max 2 x 2 / 2 208 x 208 x 64 -> 104 x 104 x 64 4 conv 128 3 x 3 / 1 104 x 104 x 64 -> 104 x 104 x 128 5 conv 64 1 x 1 / 1 104 x 104 x 128 -> 104 x 104 x 64 6 conv 128 3 x 3 / 1 104 x 104 x 64 -> 104 x 104 x 128 7 max 2 x 2 / 2 104 x 104 x 128 -> 52 x 52 x 128 8 conv 256 3 x 3 / 1 52 x 52 x 128 -> 52 x 52 x 256 9 conv 128 1 x 1 / 1 52 x 52 x 256 -> 52 x 52 x 128 10 conv 256 3 x 3 / 1 52 x 52 x 128 -> 52 x 52 x 256 11 max 2 x 2 / 2 52 x 52 x 256 -> 26 x 26 x 256 12 conv 512 3 x 3 / 1 26 x 26 x 256 -> 26 x 26 x 512 13 conv 256 1 x 1 / 1 26 x 26 x 512 -> 26 x 26 x 256 14 conv 512 3 x 3 / 1 26 x 26 x 256 -> 26 x 26 x 512 15 conv 256 1 x 1 / 1 26 x 26 x 512 -> 26 x 26 x 256 16 conv 512 3 x 3 / 1 26 x 26 x 256 -> 26 x 26 x 512 17 max 2 x 2 / 2 26 x 26 x 512 -> 13 x 13 x 512 18 conv 1024 3 x 3 / 1 13 x 13 x 512 -> 13 x 13 x1024 19 conv 512 1 x 1 / 1 13 x 13 x1024 -> 13 x 13 x 512 20 conv 1024 3 x 3 / 1 13 x 13 x 512 -> 13 x 13 x1024 21 conv 512 1 x 1 / 1 13 x 13 x1024 -> 13 x 13 x 512 22 conv 1024 3 x 3 / 1 13 x 13 x 512 -> 13 x 13 x1024 23 conv 1024 3 x 3 / 1 13 x 13 x1024 -> 13 x 13 x1024 24 conv 1024 3 x 3 / 1 13 x 13 x1024 -> 13 x 13 x1024 25 route 16 26 conv 64 1 x 1 / 1 26 x 26 x 512 -> 26 x 26 x 64 27 reorg / 2 26 x 26 x 64 -> 13 x 13 x 256 28 route 27 24 29 conv 1024 3 x 3 / 1 13 x 13 x1280 -> 13 x 13 x1024 30 conv 120 1 x 1 / 1 13 x 13 x1024 -> 13 x 13 x 120 31 detection darknet: ./src/parser.c:280: parse_region: Assertion `l.outputs == params.inputs' failed. 已放弃 (核心已转储) 这样的提示 请问这个错误提示是因为什么呢？ 另外我在安装完darknet之后按照csdn上的教程改参数，可是发现我的src文件夹中没有yolo.c等文件，但是example里面有，我就给拷贝到了scr中并做了修改，现在想想是不是安装失败了啊orz
【c-WinAPI】 程序运行崩溃（自己写的函数出错），求大神帮忙！！！！

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《程序人生》系列-这个程序员只用了20行代码就拿了冠军

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8年经验面试官详解 Java 面试秘诀
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