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？
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)