keras yolov3 tiny_yolo_body网络结构改为vgg16结构

将keras框架yolov3 tiny_yolo_body网络结构改为vgg16网络结构，程序能够运行 loss正常下降即可。

树莓派上跑yolov3-tiny.weigths出现segmentation fault是什么原因？

./darknet detector test cfg/voc.data cfg/yolov3-tiny.cfg weights/yolov3-tiny.weights data/dog.jpg 

急，跪求pycharm跑yolov3-train.py报错

 ``` import numpy as np import keras.backend as K from keras.layers import Input, Lambda from keras.models import Model from keras.callbacks import TensorBoard, ModelCheckpoint, EarlyStopping from yolo3.model import preprocess_true_boxes, yolo_body, tiny_yolo_body, yolo_loss from yolo3.utils import get_random_data def _main(): annotation_path = 'train.txt' log_dir = 'logs/000/' classes_path = 'model_data/voc_classes.txt' anchors_path = 'model_data/yolo_anchors.txt' class_names = get_classes(classes_path) anchors = get_anchors(anchors_path) input_shape = (416,416) # multiple of 32, hw model = create_model(input_shape, anchors, len(class_names) ) train(model, annotation_path, input_shape, anchors, len(class_names), log_dir=log_dir) def train(model, annotation_path, input_shape, anchors, num_classes, log_dir='logs/'): model.compile(optimizer='adam', loss={ 'yolo_loss': lambda y_true, y_pred: y_pred}) logging = TensorBoard(log_dir=log_dir) checkpoint = ModelCheckpoint(log_dir + "ep{epoch:03d}-loss{loss:.3f}-val_loss{val_loss:.3f}.h5", monitor='val_loss', save_weights_only=True, save_best_only=True, period=1) batch_size = 8 val_split = 0.1 with open(annotation_path) as f: lines = f.readlines() np.random.shuffle(lines) num_val = int(len(lines)*val_split) num_train = len(lines) - num_val print('Train on {} samples, val on {} samples, with batch size {}.'.format(num_train, num_val, batch_size)) model.fit_generator ( data_generator_wrapper ( lines[:num_train] , batch_size , input_shape , anchors , num_classes ) , steps_per_epoch=max ( 1 , num_train // batch_size ) , validation_data=data_generator_wrapper ( lines[num_train:] , batch_size , input_shape , anchors , num_classes ) , validation_steps=max ( 1 , num_val // batch_size ) , epochs=10 , initial_epoch=0 , callbacks=[logging , checkpoint] ) model.save_weights(log_dir + 'trained_weights.h5') def get_classes(classes_path): with open(classes_path) as f: class_names = f.readlines() class_names = [c.strip() for c in class_names] return class_names def get_anchors(anchors_path): with open(anchors_path) as f: anchors = f.readline() anchors = [float(x) for x in anchors.split(',')] return np.array(anchors).reshape(-1, 2) def create_model(input_shape, anchors, num_classes, load_pretrained=False, freeze_body=False, weights_path='model_data/yolo_weights.h5'): K.clear_session() # get a new session h, w = input_shape image_input = Input(shape=(w, h, 3)) num_anchors = len(anchors) y_true = [Input(shape=(h//{0:32, 1:16, 2:8}[l], w//{0:32, 1:16, 2:8}[l], num_anchors//3, num_classes+5)) for l in range(3)] model_body = yolo_body(image_input, num_anchors//3, num_classes) print('Create YOLOv3 model with {} anchors and {} classes.'.format(num_anchors, num_classes)) if load_pretrained: model_body.load_weights(weights_path, by_name=True, skip_mismatch=True) print('Load weights {}.'.format(weights_path)) if freeze_body in [1, 2]: # Do not freeze 3 output layers. num = (185 , len ( model_body.layers ) - 3)[freeze_body - 1] for i in range(num): model_body.layers[i].trainable = False print('Freeze the first {} layers of total {} layers.'.format(num, len(model_body.layers))) model_loss = Lambda ( yolo_loss , output_shape=(1 ,) , name='yolo_loss', arguments={'anchors': anchors, 'num_classes': num_classes, 'ignore_thresh': 0.5} )(model_body.output + y_true) model = Model(inputs=[model_body.input] + y_true, outputs=model_loss) return model def data_generator(annotation_lines, batch_size, input_shape, anchors, num_classes): n = len(annotation_lines) i = 0 while True: image_data = [] box_data = [] for b in range(batch_size): if i==0: np.random.shuffle(annotation_lines) image, box = get_random_data(annotation_lines[i], input_shape, random=True) image_data.append(image) box_data.append(box) i = (i+1) % n image_data = np.array(image_data) box_data = np.array(box_data) y_true = preprocess_true_boxes(box_data, input_shape, anchors, num_classes) yield [image_data]+y_true, np.zeros(batch_size) def data_generator_wrapper(annotation_lines, batch_size, input_shape, anchors, num_classes): n = len(annotation_lines) if n==0 or batch_size<=0: return None return data_generator(annotation_lines, batch_size, input_shape, anchors, num_classes) if __name__ == '__main__': _main() ``` 报了一个:tensorflow.python.framework.errors_impl.InvalidArgumentError: Inputs to operation training/Adam/gradients/AddN_24 of type _MklAddN must have the same size and shape. Input 0: [2768896] != input 1: [8,26,26,512] [[Node: training/Adam/gradients/AddN_24 = _MklAddN[N=2, T=DT_FLOAT, _kernel="MklOp", _device="/job:localhost/replica:0/task:0/device:CPU:0"](training/Adam/gradients/batch_normalization_65/FusedBatchNorm_grad/FusedBatchNormGrad, training/Adam/gradients/batch_n

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

Problem Description The tiny country of Waterlogged is protected by a series of levees that form a quadrilateral as shown below: The quadrilateral is defined by four vertices. The levees partition the country into four quadrants. Each quadrant is identified by a pair of vertices representing the outside edge of that quadrant. For example, Quadrant 1 shown below is defined by the points (x1, y1) and (x2, y2) . It happens very often that the country of Waterlogged becomes flooded, and the levees need to be reinforced, but their country is poor and they have limited resources. They would like to be able to reinforce those levees that encompass the largest area first, then the next largest second, then the next largest third, and the smallest area fourth. Help Waterlogged identify which quadrants are the largest, and the length of the levees around them. Input here will be several sets of input. Each set will consist of eight real numbers, on a single line. Those numbers will represent, in order: X1 Y1 X2 Y2 X3 Y3 X4 Y4 The four points are guaranteed to form a convex quadrilateral when taken in order -- that is, there will be no concavities, and no lines crossing. Every number will be in the range from -1000.0 to 1000.0 inclusive. No Quadrant will have an area or a perimeter smaller than 0.001. End of the input will be a line with eight 0.0's. Output For each input set, print a single line with eight floating point numbers. These represent the areas and perimeters of the four Quadrants, like this: A1 P1 A2 P2 A3 P3 A4 P4 Print them in order from largest area to smallest -- so A1 is the largest area. If two Quadrants have the same area when rounded to 3 decimal places, output the one with the largest perimeter first. Print all values with 3 decimal places of precision (rounded). Print spaces between numbers. Do not print any blank lines between outputs. Sample Input 1 2 1 5 5 2 2 0 3.5 2.2 4.8 -9.6 -1.2 -4.4 -8.9 12.4 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 Sample Output 5.100 11.459 3.400 9.045 0.900 6.659 0.600 4.876 44.548 38.972 21.982 25.997 20.342 38.374 10.038 19.043

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？

字符串代码处理，具体怎么写，Ye Holy Hand Grenades!

Problem Description Holy Hand Grenades have found multiple uses throughout history. Most notably, King Arthur did use a Holy Hand Grenade (HHG) to dispatch a vicious rabbit which was guarding the entrance to a cave. The instructions for its use are in the Book of Armaments (Chapter 2, verses 9-21), and are read as follows: Unfortunately, the Book of Armaments faileth to mention that the HHG must come to a complete stop (i.e. rest stably) before it can explodeth and blow a rabbit to tiny bits. As long as it rolleth or wobbeleth, it refuseth to explodeth. Unknown is the reason why the authors of the Book of Armaments neglected to mention this pertinent fact. Further complicating matters, different HHGs have different radii of destruction (RD). Once a HHG resteth stably, and explodeth, everything within or equal to its RD shalt be snuffed out. Presumably, this is what happened to the rabbit with big, pointy teeth; however, there remaineth some controversy over this matter. Because of this controversy, the king hath decreed a simulation to be run that may possibly answer this question. Your task is to determine the result of the grenade’s explosion. The grenade must always be at rest in order to detonate. All HHGs have a radius R=1 and a variable radius of destruction (RD). At the instant of detonation, the rabbit could be mid-leap, on the ground, or underground. For each simulation the terrain remaineth unchanged. It consisteth of two intersecting curves, y = e-x and y = ln(x) as we have revealed unto you in the illustrious illustration. The curves reside in the plane formed by the rabbit, the grenade, and the center of the earth. The terrain is impervious to detonations, and changeth not between tests. If a bunny’s center of mass is strictly below the ground level (as denoted by the intersecting curves) when a HHG explodeth, that bunny shalt remaineth un-snuffedeth and thus shalt live to bite another day. If the bunny resideth strictly outside of the range of a stably resting HHG’s radius of destruction, that bunny shalt also remain un-snuffedeth out. Each result shalt be dependent upon computations accurate even unto 8 significant figures. There shalt not be an input value smaller even than 1.0E-15 or greater even than 1.0E+15. Because the bunny’s center of mass doth be its location, a bunny might possibly end up inside both the RD and the R=1 of a HHG. This simply means the bunny is curled around the grenade. If the center of mass of such a bunny be strictly inside the RD it shall most surely be snufféd out. Amen. . Input The first line of input doth contain yea verily a single integer indicating the number of holy hand grenades in the data file. Each line of the file doth contain the radius of destruction of this particular holy hand grenade, and finally doth contain the rabbit’s x coordinate (xb) and y coordinate (yb) during this particular test run. Output For each test run, thy program shalt print out the words "Bunny Bits" (if the rabbit is blown to bits by the grenade) or "Bunny Biteth Knights" (if the rabbit liveth to fight on another day anon). Sample Input 2 1.5 3.0 3.0 5.5 3.0 3.0 Sample Output Bunny Biteth Knights Bunny Bits

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.

友善之臂tiny4412ADK 1312开发板通过gpio口控制外接led灯问题

最近自己在用友善的tiny4412ADK 1312开发板，目的是想通过gpio口来控制外接led灯。目前自己可以实现资料上所给的GPIO_LED_Demo程序例子,点亮核心板上的led1~4灯，但对于外接led灯这个过程不是太了解，以及如何通过gpio口控制外接led灯？希望有这方面经验的前辈给我点建议，谢谢！

激光的程序怎么设计Laser Beam

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

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