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#include<iom16v.h>
#include<macros.h>
#define uint unsigned int
#define uchar unsigned char
uint a,b,c,d=0;
void delay(c)
{ for for(a=0;a<c;a++)
for(b=0;b<12;b++);
};
uchar tab[]={
0xc0,0xf9,0xa4,0xb0,0x99,0x92,0x82,0xf8,0x80,0x90,
標簽：
AVR
單片機
數碼管
上傳時間：
2013-10-21
上傳用戶：13788529953
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三種方法讀取鍵值 使用者設計行列鍵盤介面，一般常採用三種方法讀取鍵值。 中斷式 在鍵盤按下時產生一個外部中斷通知CPU，並由中斷處理程式通過不同位址讀資料線上的狀態判斷哪個按鍵被按下。 本實驗採用中斷式實現使用者鍵盤介面。 掃描法 對鍵盤上的某一行送低電位，其他為高電位，然後讀取列值，若列值中有一位是低，表明該行與低電位對應列的鍵被按下。否則掃描下一行。 反轉法 先將所有行掃描線輸出低電位，讀列值，若列值有一位是低表明有鍵按下；接著所有列掃描線輸出低電位，再讀行值。 根據讀到的值組合就可以查表得到鍵碼。4x4鍵盤按4行4列組成如圖電路結構。按鍵按下將會使行列連成通路，這也是見的使用者鍵盤設計電路。
//-----------4X4鍵盤程序--------------// uchar keboard(void) { uchar xxa,yyb,i,key; if((PINC&0x0f)!=0x0f) //是否有按鍵按下 {delayms(1); //延時去抖動 if((PINC&0x0f)!=0x0f) //有按下則判斷 { xxa=~(PINC|0xf0); //0000xxxx DDRC=0x0f; PORTC=0xf0; delay_1ms(); yyb=~(PINC|0x0f); //xxxx0000 DDRC=0xf0; //復位 PORTC=0x0f; while((PINC&0x0f)!=0x0f) //按鍵是否放開 { display(data); } i=4; //計算返回碼 while(xxa!=0) { xxa=xxa>>1; i--; } if(yyb==0x80) key=i; else if(yyb==0x40) key=4+i; else if(yyb==0x20) key=8+i; else if(yyb==0x10) key=12+i; return key; //返回按下的鍵盤碼 } } else return 17; //沒有按鍵按下 }
標簽：
4x4
鍵盤
上傳時間：
2013-11-12
上傳用戶：a673761058
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Single-Ended and Differential S-Parameters
Differential circuits have been important incommunication systems for many years. In the past,differential communication circuits operated at lowfrequencies, where they could be designed andanalyzed using lumped-element models andtechniques. With the frequency of operationincreasing beyond 1GHz, and above 1Gbps fordigital communications, this lumped-elementapproach is no longer valid, because the physicalsize of the circuit approaches the size of awavelength.Distributed models and analysis techniques are nowused instead of lumped-element techniques.Scattering parameters, or S-parameters, have beendeveloped for this purpose [1]. These S-parametersare defined for single-ended networks. S-parameterscan be used to describe differential networks, but astrict definition was not developed until Bockelmanand others addressed this issue [2]. Bockelman’swork also included a study on how to adapt single-ended S-parameters for use with differential circuits[2]. This adaptation, called “mixed-mode S-parameters,” addresses differential and common-mode operation, as well as the conversion betweenthe two modes of operation.This application note will explain the use of single-ended and mixed-mode S-parameters, and the basicconcepts of microwave measurement calibration.
標簽：
差分電路
單端
模式
上傳時間：
2014-03-25
上傳用戶：yyyyyyyyyy
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電路板故障分析
維修方式介紹
ASA維修技術
ICT維修技術
沒有線路圖,無從修起
電路板太複雜,維修困難
維修經驗及技術不足
無法維修的死板,廢棄可惜
送電中作動態維修,危險性極高
備份板太多,積壓資金
送國外維修費用高,維修時間長
對老化零件無從查起無法預先更換
維修速度及效率無法提升,造成公司負擔,客戶埋怨
投資大量維修設備,操作複雜,績效不彰
標簽：
電路板維修
技術資料
上傳時間：
2013-11-09
上傳用戶：chengxin
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題目：利用條件運算符的嵌套來完成此題：學習成績>=90分的同學用A表示，60-89分之間的用B表示，60分以下的用C表示。 1.程序分析：(a>b)?a:b這是條件運算符的基本例子。
標簽：
gt
90
運算符
嵌套
上傳時間：
2015-01-08
上傳用戶：lifangyuan12
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RSA算法 :首先, 找出三個數, p, q, r, 其中 p, q 是兩個相異的質數, r 是與 (p-1)(q-1) 互質的數...... p, q, r 這三個數便是 person_key，接著, 找出 m, 使得 r^m == 1 mod (p-1)(q-1)..... 這個 m 一定存在, 因為 r 與 (p-1)(q-1) 互質, 用輾轉相除法就可以得到了..... 再來, 計算 n = pq....... m, n 這兩個數便是 public_key ，編碼過程是, 若資料為 a, 將其看成是一個大整數, 假設 a < n.... 如果 a >= n 的話, 就將 a 表成 s 進位 (s
標簽：
person_key
RSA
算法
上傳時間：
2013-12-14
上傳用戶：zhuyibin
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一個LCD燈的小程序。不是我寫的。我只負責了調試。適用在ACEXEP1K30QC208-3上。我跑了SIMULATOR，管腳連接標示了。我也下在電路板上試過了，沒有問題。要用到實驗板上的兄弟們把CLK1改到TESTOUT3或者0就好了。綫幫助新手，人人有責。
標簽：
SIMULATOR
ACEXEP
LCD
208
上傳時間：
2015-04-10
上傳用戶：330402686
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數字運算，判斷一個數是否接近素數
A Niven number is a number such that the sum of its digits divides itself. For example, 111 is a Niven number because the sum of its digits is 3, which divides 111. We can also specify a number in another base b, and a number in base b is a Niven number if the sum of its digits divides its value.
Given b (2 <= b <= 10) and a number in base b, determine whether it is a Niven number or not.
Input
Each line of input contains the base b, followed by a string of digits representing a positive integer in that base. There are no leading zeroes. The input is terminated by a line consisting of 0 alone.
Output
For each case, print "yes" on a line if the given number is a Niven number, and "no" otherwise.
Sample Input
10 111
2 110
10 123
6 1000
8 2314
0
Sample Output
yes
yes
no
yes
no
標簽：
數字
運算
上傳時間：
2015-05-21
上傳用戶：daguda
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源代碼\用動態規劃算法計算序列關系個數
用關系"<"和"="將3個數a，b，c依次序排列時，有13種不同的序列關系：
a=b=c,a=b<c,a<b=v,a<b<c,a<c<b
a=c<b,b<a=c,b<a<c,b<c<a,b=c<a
c<a=b,c<a<b,c<b<a
若要將n個數依序列，設計一個動態規劃算法，計算出有多少種不同的序列關系，
要求算法只占用O(n),只耗時O(n*n).
標簽：
lt
源代碼
動態規劃
序列
上傳時間：
2013-12-26
上傳用戶：siguazgb
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The government of a small but important country has decided that the alphabet needs to be streamlined and reordered. Uppercase letters will be eliminated. They will issue a royal decree in the form of a String of B and A characters. The first character in the decree specifies whether a must come ( B )Before b in the new alphabet or ( A )After b . The second character determines the relative placement of b and c , etc. So, for example, "BAA" means that a must come Before b , b must come After c , and c must come After d .
Any letters beyond these requirements are to be excluded, so if the decree specifies k comparisons then the new alphabet will contain the first k+1 lowercase letters of the current alphabet.
Create a class Alphabet that contains the method choices that takes the decree as input and returns the number of possible new alphabets that conform to the decree. If more than 1,000,000,000 are possible, return -1.
Definition
標簽：
government
streamline
important
alphabet
上傳時間：
2015-06-09
上傳用戶：weixiao99
