This example streams input from a ADC source to a DAC. An analog signal is acquired block-by-block into SDRAM from the ADC (an AD9244 in this example). The frames are then output with a one-frame delay to the DAC (an AD9744 in this example). In this example, no processing is done on the frames. They are passed unaltered.
標(biāo)簽： block-by-block acquired example streams
上傳時(shí)間： 2015-12-29
上傳用戶：bjgaofei
~{JGR 8vQ IzWwR5SC5D2V?bD#DbO5M3~} ~{3v?b~} ~{Hk?b~} ~{2iQ/5H9&D\~} ~{?IRTWw@)3d~} ~{TZ~}JDK1.4.2~{OBM(9}~}
上傳時(shí)間： 2015-02-22
上傳用戶：ommshaggar
b to b 模式 電子商務(wù)系統(tǒng) ，c# 開發(fā) , B/S結(jié)構(gòu)
標(biāo)簽： to 模式 電子商務(wù)系統(tǒng)
上傳時(shí)間： 2014-01-20
上傳用戶：hanli8870
a XOR b> a,然后a XOR b< b,and both a and b are dependent data
標(biāo)簽： XOR and dependent both
上傳時(shí)間： 2014-01-27
上傳用戶：yxgi5
樣板 B 樹 ( B - tree ) 規(guī)則 : (1) 每個(gè)節(jié)點(diǎn)內(nèi)元素個(gè)數(shù)在 [MIN,2*MIN] 之間, 但根節(jié)點(diǎn)元素個(gè)數(shù)為 [1,2*MIN] (2) 節(jié)點(diǎn)內(nèi)元素由小排到大, 元素不重複 (3) 每個(gè)節(jié)點(diǎn)內(nèi)的指標(biāo)個(gè)數(shù)為元素個(gè)數(shù)加一 (4) 第 i 個(gè)指標(biāo)所指向的子節(jié)點(diǎn)內(nèi)的所有元素值皆小於父節(jié)點(diǎn)的第 i 個(gè)元素 (5) B 樹內(nèi)的所有末端節(jié)點(diǎn)深度一樣
上傳時(shí)間： 2017-05-14
上傳用戶：日光微瀾
歐幾里德算法：輾轉(zhuǎn)求余 原理: gcd(a,b)=gcd(b,a mod b) 當(dāng)b為0時(shí),兩數(shù)的最大公約數(shù)即為a getchar()會(huì)接受前一個(gè)scanf的回車符
標(biāo)簽： gcd getchar scanf mod
上傳時(shí)間： 2014-01-10
上傳用戶：2467478207
Algorithm Festel. Programs read from file to array byte[]. key block = 64 data block = 128 if use classis Festel data block = 256 if use modification Festel
標(biāo)簽： block Algorithm Programs Festel
上傳時(shí)間： 2014-12-05
上傳用戶：kikye
數(shù)據(jù)結(jié)構(gòu)課程設(shè)計(jì) 數(shù)據(jù)結(jié)構(gòu)B+樹 B+ tree Library
標(biāo)簽： Library tree 數(shù)據(jù)結(jié)構(gòu) 樹
上傳時(shí)間： 2013-12-31
上傳用戶：semi1981
* 高斯列主元素消去法求解矩陣方程AX=B,其中A是N*N的矩陣,B是N*M矩陣 * 輸入: n----方陣A的行數(shù) * a----矩陣A * m----矩陣B的列數(shù) * b----矩陣B * 輸出: det----矩陣A的行列式值 * a----A消元后的上三角矩陣 * b----矩陣方程的解X
上傳時(shí)間： 2015-07-26
上傳用戶：xauthu
We have a group of N items (represented by integers from 1 to N), and we know that there is some total order defined for these items. You may assume that no two elements will be equal (for all a, b: a<b or b<a). However, it is expensive to compare two items. Your task is to make a number of comparisons, and then output the sorted order. The cost of determining if a < b is given by the bth integer of element a of costs (space delimited), which is the same as the ath integer of element b. Naturally, you will be judged on the total cost of the comparisons you make before outputting the sorted order. If your order is incorrect, you will receive a 0. Otherwise, your score will be opt/cost, where opt is the best cost anyone has achieved and cost is the total cost of the comparisons you make (so your score for a test case will be between 0 and 1). Your score for the problem will simply be the sum of your scores for the individual test cases.
標(biāo)簽： represented integers group items
上傳時(shí)間： 2016-01-17
上傳用戶：jeffery
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