启航电商ERP系统是一套为中小电商企业构建的一套简单、实用、覆盖全流程的电商系统,本项目采用SpringBoot+Vue2前后端分离开发。 支持供应商一件代发和仓库发货两种发货方式,功能覆盖采购、网店订单处理、供应商一件代发、仓库发货、网店售后、网店商品管理、仓库出入库、采购结算
Qihang E-Commerce ERP system is a simple, practical and full process e-commerce system built for small and medium-sized e-commerce enterprises. The project adopts SpringBoot+Vue2 to separate the front and rear ends. Two delivery methods are supported: supplier one-piece shipment and warehouse shipment. The functions cover purchase, online store order processing, supplier one-piece shipment, warehouse shipment, online store after-sales, online store commodity management, warehouse issue/receipt, and purchase settlement (2024-03-13, Java, 0KB, 下载0次)
题目：企业发放的奖金根据利润提成。利润(I)低于或等于10万元时，奖金可提10 ；利润高 于10万元，低于20万元时，低于10万元的部分按10 提成，高于10万元的部分，可可提 成7.5 ；20万到40万之间时，高于20万元的部分，可提成5 ；40万到60万之间时高于 40万元的部分，可提成3 ；60万到100万之间时，高于60万元的部分，可提成1.5 ，高于 100万元时，超过100万元的部分按1 提成，从键盘输入当月利润I，求应发放奖金总数？
Topic: enterprise extending bonus profits according to the commission. Profits (I) is less than or equal to 100000 yuan, the prize can carry 10 High profits
In 100000 yuan, less than 200000 yuan, is below 100000 yuan part according to 10 of commission, part of the higher than 100000 yuan, cocoa mention
Into 7.5 Between 200000 and 400000, more than 200000 yuan part, can commission 5 Between 400000 and 600000 is higher than the
Part of the 400000 yuan, can commission 3 Between 600000 and 1 million, more than 600000 yuan part, can commission 1.5 , higher than
1 million yuan, more than 1 million yuan part according to 1 commission from the keyboard that month profit I, for total should be bonuses?
(2012-05-28, Visual C++, 2KB, 下载1次)
企业发放的奖金根据利润提成。利润(I)低于或等于10万元时，奖金可提10 ；利润高
于10万元，低于20万元时，低于10万元的部分按10 提成，高于10万元的部分，可可提
成7.5 ；20万到40万之间时，高于20万元的部分，可提成5 ；40万到60万之间时高于
40万元的部分，可提成3 ；60万到100万之间时，高于60万元的部分，可提成1.5 ，高于
100万元时，超过100万元的部分按1 提成，从键盘输入当月利润I，求应发放奖金总数？
Company bonus according to the percentage of profits. Profit ( I ) less than or equal to 100000yuan, the bonus is10 high profits
In100000 yuan, less than 200000 yuan, less than $100000as part of the10 Commission, more than 100000 yuan, cocoa.
7.5 200000to 400000, more than 200000 yuan, can generate 5 400000to 600000is higher than the
400000part, can generate 3 600000to 1000000, more than 600000 yuan, but percentage of 1.5 , higher than the
1000000 yuan, more than 1000000 yuan as part of the1 Commission, from the keyboard input of the I should seek profit, total bonus? (2012-04-25, C/C++, 3KB, 下载2次)
企业发放的奖金根据利润提成。利润(I)低于或等于10万元时，奖金可提10 ；利润高
于10万元，低于20万元时，低于10万元的部分按10 提成，高于10万元的部分，可可提
成7.5 ；20万到40万之间时，高于20万元的部分，可提成5 ；40万到60万之间时高于
40万元的部分，可提成3 ；60万到100万之间时，高于60万元的部分，可提成1.5 ，高于
100万元时，超过100万元的部分按1 提成，从键盘输入当月利润I，求应发放奖金总数？
Company bonus according to the percentage of profits. Profit ( I ) less than or equal to 100000yuan, the bonus is10 high profits
In100000 yuan, less than 200000 yuan, less than $100000as part of the10 Commission, more than 100000 yuan, cocoa.
7.5 200000to 400000, more than 200000 yuan, can generate 5 400000to 600000is higher than the
400000part, can generate 3 600000to 1000000, more than 600000 yuan, but percentage of 1.5 , higher than the
1000000 yuan, more than 1000000 yuan as part of the1 Commission, from the keyboard input of the I should seek profit, total bonus? (2012-04-25, C/C++, 3KB, 下载2次)
企业发放的奖金根据利润提成。利润(I)低于或等于10万元时，奖金可提10 ；利润高
于10万元，低于20万元时，低于10万元的部分按10 提成，高于10万元的部分，可可提
成7.5 ；20万到40万之间时，高于20万元的部分，可提成5 ；40万到60万之间时高于
40万元的部分，可提成3 ；60万到100万之间时，高于60万元的部分，可提成1.5 ，高于
100万元时，超过100万元的部分按1 提成，从键盘输入当月利润I，求应发放奖金总数？
Corporate bonuses based on profit commission. Profit (I) is less than or equal to 10 million, the bonus can mention 10 high profits
$ 100,000, less than 20 million, less than 10 million part of the 10 commission, higher than 10 million part of the cocoa mention
Into 7.5 200000-400000, more than 20 million part of the commission 5 between 40 million to 60 million higher than the
40 million part of the commission 3 between 60 million to 100 million, more than 60 million part, to the commission of 1.5 , higher than the
1 million yuan, more than 100 million parts of 1 commission from the keyboard, enter the month profits I seek the total number of bonuses? (2012-02-13, C/C++, 1KB, 下载2次)
针对多径效应的影响，提出了一种基于矩阵束的MIMO 雷达低仰角快速估计方法。该方法同时考虑了发射多径信号和接收多径信号，采用单样本数信号矢量构造了一个前后向矩阵束，并利用两个酉矩阵对该矩阵束进行降维处理，最后采用广义特征值分解的总体最小二乘法来估计目标角度。算法不需要估计数据协方差矩阵，可在低
信噪比和单样本数情况下，有效地克服多径效应，实现同时多目标低仰角估计，相比最大似然算法，避免了谱峰搜索，计算量小。仿真结果验证了该算法的有效性。
To overcome the multipath effect, a fast algorithm for low elevation estimation for MIMO radar via matrix pencil is proposed. The signals both in transmit multipath and receive multipath are considered. Firstly, a forward-backward matrix pencil is formed based on the single-sampled vector. Secondly, the reduced-dimensional processing is applied to the matrix pencil via two unitary matrices. Finally, the generalized eigen-decomposition is employed to estimate the low elevation in multipath environments directly. Without the covariance matrix estimation, the proposed method can overcome multipath effect effectively in case of low SNR and estimate multi-target simultaneously. Compared with Maximum Likelihood (ML) method, it avoids spectrum peak searching and reduces the computational burden. Numerical results verify the effectiveness of this method (2011-09-03, Others, 249KB, 下载183次)
在一个按照东西和南北方向划分成规整街区的城市里，n 个居民点散乱地分布在不同的街区中。用x 坐标表示东西向，用y 坐标表示南北向。各居民点的位置可以由坐标(x,y)表示。街区中任意2 点(x1, y1) 和(x2, y2) 之间的距离可以用数值| x1 . x2| + | y1 . y2| 度量。
居民们希望在城市中选择建立邮局的最佳位置，使n 个居民点到邮局的距离总和最小。
In a direction in accordance with the East-West and North-South divided into regular city blocks, n and the distribution of settlements scattered in different neighborhoods in. Things used to express the coordinates x, y coordinates used to express the North and the South. The location of the settlements by the coordinates (x, y) express. Arbitrary blocks 2:00 (x1, y1) and (x2, y2) can use the distance between the values | x1. X2 |+ | Y1. Y2 | metrics. Residents want the city to select the best location to establish post offices, so that n settlements to the post office the sum of the smallest distance. (2008-07-07, Visual C++, 1KB, 下载9次)
C-N解偏微分方程的程序。解的是定步长的抛物型偏微分方程：
du/dx - a * d2u/dx2 = 0
在程序中可以更改 a 的值以实现不同系数的解。在循环中改变 a 的值以实现变系数. 该说明在压缩包是也有
CN solution of partial differential equations procedures. Solution is scheduled to step parabolic partial differential equation: du/dx- a* d2u/dx2 = 0 in the procedure can be changed in order to realize the value of a different coefficient of the solution. Change in the cycle in order to realize the value of a variable coefficient. The note in the compressed packet is also (2008-04-20, matlab, 1KB, 下载236次)
高分辨率要求系统具有大的带宽，瞬时带宽的增加必将提高系统对硬件的要求，本文采用方便灵活的步进频率波形信号。脉间频率步进波形通过子脉冲载频的步进变化来获得大的有效带宽，使成像具有高分辨率，采用加窗和补零方法提高信噪比，但该信号对目标径向速度非常敏感。采用补零方法提高距离取样分辨率，使距离像细化，并用公式说明了补零只能提高距离取样分辨率，并不能改变频率步进信号的距离分辨能力。该信号波形对目标径向速度的敏感，使目标能量分散到邻近的距离单元造成距离分辨率下降，如果不事先进行速度补偿，直接对回波信号进行逆傅立叶变换，将使所成一维距离像发生频谱展宽，并伴有距离像发生平移。本文证明了当目标有径向速度时仍采用对回波信号直接进行逆傅立叶变换的方法将使一维距离像发生频谱展宽并伴有距离像发生频移，从而严重影响了一维距离像的质量。
high-resolution requirements of the system with large bandwidth, the increased instantaneous bandwidth will improve the system of hardware, In this paper, a flexible and convenient step frequency signals. Pulse frequency step-pulse waveform through the carrier frequency step to achieve great changes in the effective bandwidth. make imaging with high resolution, increasing the window and fill methods to improve signal-to-noise ratio is, However, the signal on the target radial velocity is very sensitive. Using zero-distance sampling method to improve resolution, so that distance as refinement, and formulates a zero-distance sampling can improve resolution, does not alter the frequency of the signal from stepping resolution capabilities. The signal waveform on the target radial velocity sens (2007-04-07, matlab, 2KB, 下载472次)