哈夫曼樹又稱最優二叉樹,是一種帶權路徑長度最短的二叉樹。所謂樹的帶權路徑長度,就是樹中所有的葉結點的權值乘上其到根結點的路徑長度(若根結點為0層,葉結點到根結點的路徑長度為葉結點的層數)。樹的帶權路徑長度記為WPL=(W1*L1+W2*L2+W3*L3+...+Wn*Ln),N個權值Wi(i=1,2,...n)構成一棵有N個葉結點的二叉樹,相應的葉結點的路徑長度為Li(i=1,2,...n)。可以證明哈夫曼樹的WPL是最小的。
上傳時間: 2017-06-09
上傳用戶:wang5829
g a w k或GNU awk是由Alfred V. A h o,Peter J.We i n b e rg e r和Brian W. K e r n i g h a n于1 9 7 7年為U N I X創建的a w k編程語言的較新版本之一。a w k出自創建者姓的首字母。a w k語言(在其所有的版本中)是一種具有很強能力的模式匹配和過程語言。a w k獲取一個文件(或多個文件)來查找匹配特定模式的記錄。當查到匹配后,即執行所指定的動作。作為一個程序員,你不必操心通過文件打開、循環讀每個記錄,控制文件的結束,或執行完后關閉文件。
上傳時間: 2014-01-02
上傳用戶:hwl453472107
OFDM程序,這么安排矩陣的目的是為了構造共軛對稱矩陣 共軛對稱矩陣的特點是 在ifft/fft的矢量上 N點的矢量 在0,N/2點必須是實數 一般選為0 1至N/2點 與 (N/2)+1至N-1點關于N/2共軛對稱
上傳時間: 2014-02-05
上傳用戶:woshiayin
在對一些變步長LMS算法分析的基礎上,提出了步長因子 (n)與誤差信號e(n)之間一種新的非線性函數關系
上傳時間: 2014-01-16
上傳用戶:LIKE
FFT,即為快速傅氏變換,是離散傅氏變換的快速算法,它是根據離散傅氏變換的奇、偶、虛、實等特性,對離散傅立葉變換的算法進行改進獲得的。它對傅氏變換的理論并沒有新的發現,但是對于在計算機系統或者說數字系統中應用離散傅立葉變換,可以說是進了一大步。 設x(n)為N項的復數序列,由DFT變換,任一X(m)的計算都需要N次復數乘法和N-1次復數加法,而一次復數乘法等于四次實數乘法和兩次實數加法,
標簽: FFT
上傳時間: 2013-12-29
上傳用戶:13517191407
RSA ( Rivest Shamir Adleman )is crypthograph system that used to give a secret information and digital signature . Its security based on Integer Factorization Problem (IFP). RSA uses an asymetric key. RSA was created by Rivest, Shamir, and Adleman in 1977. Every user have a pair of key, public key and private key. Public key (e) . You may choose any number for e with these requirements, 1< e <Æ (n), where Æ (n)= (p-1) (q-1) ( p and q are first-rate), gcd (e,Æ (n))=1 (gcd= greatest common divisor). Private key (d). d=(1/e) mod(Æ (n)) Encyption (C) . C=Mª mod(n), a = e (public key), n=pq Descryption (D) . D=C° mod(n), o = d (private key
標簽: crypthograph information Adleman Rivest
上傳時間: 2017-09-01
上傳用戶:chfanjiang
某市進行了一次英語競賽,一共有N個人報名參加(N<1000000),按照報名順序分別是1號~N號。英語競賽的得分范圍是0~100分,可能會有0.5分存在。現在給你N的值和按照號碼排列的得分,要求輸出第一名的號碼。若是有并列第一名則需要同屬輸出多個第一名的號碼。
標簽: C語言代碼
上傳時間: 2016-03-29
上傳用戶:我是鑫鑫
輸入N*N(N<100)的矩陣,輸出它的轉置矩陣。 第一行為整數N。 接著是一個N*N的矩陣
標簽: 矩陣轉置
上傳時間: 2016-03-29
上傳用戶:我是鑫鑫
簡單命令使用grep等的使用 [zorro@isch ~]$ history 1 ifconfig 2 su 3 exit 4 ls 5 cd Desktop/ 6 ls 7 tar zxcf VMwareTools-8.4.5-324285.tar.gz 8 tar zxvf VMwareTools-8.4.5-324285.tar.gz 9 cd vmware-tools-distrib/ 10 ls 11 ./vmware-install.pl 12 su 13 ls 14 cd .. 15 ls 16 rm VMwareTools-8.4.5-324285.tar.gz 17 rm -r vmware-tools-distrib 18 ls 19 make 20 ls 21 cd redis/ 22 quit 23 ls 24 ca redis/ 25 cd redis/ 26 cd redis-2.8.17 27 make 28 cd redis-2.8.17 29 ls 30 cd redis-2.8.17 31 cd str 32 cd src 33 ls 34 ./redis-cli 35 ls 36 cd redis-2.8.17 tar.gz 37 make 38 cd src 39 ./redis-server .. /redis.conf 40 ./redis-cli 41 ./redis-server ../redis.conf 42 vi test1.sh 43 ./test1.sh 44 vi test.sh 45 ./test.sh 46 ls 47 chmod 777 test.sh 48 ./test.sh 49 vi express 50 $ grep –n ‘the’ express 51 clear 52 grep -n 'the' express 53 vi express 54 grep -n 'the' express 55 grep -vn 'the'express 56 grep -vn 'the' express 57 grep -in 'the' express 58 vi test2.c 59 grep -l 'the' *.c 60 grep -n 't[ae]st' express 61 grep -n 'oo' express 62 grep -n '[^g]oo' express 63 grep -n '[a^z]oo' express 64 grep -n '[0^9]' express 65 grep -n '^the' express 66 vi express 67 sed -e 'd' express 68 sed -e '1d' express 69 sed -e '1~7d' express 70 sed -e '$d' express 71 sed -e '1,/^$/d' express 72 ls 73 cd 74 pwd 75 history [zorro@isch ~]$
標簽: 簡單命令使用
上傳時間: 2016-05-24
上傳用戶:12345678gan
#include <malloc.h> #include <stdio.h> #include <stdlib.h> #include <string.h> #define NULL 0 #define MaxSize 30 typedef struct athletestruct /*運動員*/ { char name[20]; int score; /*分數*/ int range; /**/ int item; /*項目*/ }ATH; typedef struct schoolstruct /*學校*/ { int count; /*編號*/ int serial; /**/ int menscore; /*男選手分數*/ int womenscore; /*女選手分數*/ int totalscore; /*總分*/ ATH athlete[MaxSize]; /**/ struct schoolstruct *next; }SCH; int nsc,msp,wsp; int ntsp; int i,j; int overgame; int serial,range; int n; SCH *head,*pfirst,*psecond; int *phead=NULL,*pafirst=NULL,*pasecond=NULL; void create(); void input () { char answer; head = (SCH *)malloc(sizeof(SCH)); /**/ head->next = NULL; pfirst = head; answer = 'y'; while ( answer == 'y' ) { Is_Game_DoMain: printf("\nGET Top 5 when odd\nGET Top 3 when even"); printf("\n輸入運動項目序號 (x<=%d):",ntsp); scanf("%d",pafirst); overgame = *pafirst; if ( pafirst != phead ) { for ( pasecond = phead ; pasecond < pafirst ; pasecond ++ ) { if ( overgame == *pasecond ) { printf("\n這個項目已經存在請選擇其他的數字\n"); goto Is_Game_DoMain; } } } pafirst = pafirst + 1; if ( overgame > ntsp ) { printf("\n項目不存在"); printf("\n請重新輸入"); goto Is_Game_DoMain; } switch ( overgame%2 ) { case 0: n = 3;break; case 1: n = 5;break; } for ( i = 1 ; i <= n ; i++ ) { Is_Serial_DoMain: printf("\n輸入序號 of the NO.%d (0<x<=%d): ",i,nsc); scanf("%d",&serial); if ( serial > nsc ) { printf("\n超過學校數目,請重新輸入"); goto Is_Serial_DoMain; } if ( head->next == NULL ) { create(); } psecond = head->next ; while ( psecond != NULL ) { if ( psecond->serial == serial ) { pfirst = psecond; pfirst->count = pfirst->count + 1; goto Store_Data; } else { psecond = psecond->next; } } create(); Store_Data: pfirst->athlete[pfirst->count].item = overgame; pfirst->athlete[pfirst->count].range = i; pfirst->serial = serial; printf("Input name:) : "); scanf("%s",pfirst->athlete[pfirst->count].name); } printf("\n繼續輸入運動項目(y&n)?"); answer = getchar(); printf("\n"); } } void calculate() /**/ { pfirst = head->next; while ( pfirst->next != NULL ) { for (i=1;i<=pfirst->count;i++) { if ( pfirst->athlete[i].item % 2 == 0 ) { switch (pfirst->athlete[i].range) { case 1:pfirst->athlete[i].score = 5;break; case 2:pfirst->athlete[i].score = 3;break; case 3:pfirst->athlete[i].score = 2;break; } } else { switch (pfirst->athlete[i].range) { case 1:pfirst->athlete[i].score = 7;break; case 2:pfirst->athlete[i].score = 5;break; case 3:pfirst->athlete[i].score = 3;break; case 4:pfirst->athlete[i].score = 2;break; case 5:pfirst->athlete[i].score = 1;break; } } if ( pfirst->athlete[i].item <=msp ) { pfirst->menscore = pfirst->menscore + pfirst->athlete[i].score; } else { pfirst->womenscore = pfirst->womenscore + pfirst->athlete[i].score; } } pfirst->totalscore = pfirst->menscore + pfirst->womenscore; pfirst = pfirst->next; } } void output() { pfirst = head->next; psecond = head->next; while ( pfirst->next != NULL ) { // clrscr(); printf("\n第%d號學校的結果成績:",pfirst->serial); printf("\n\n項目的數目\t學校的名字\t分數"); for (i=1;i<=ntsp;i++) { for (j=1;j<=pfirst->count;j++) { if ( pfirst->athlete[j].item == i ) { printf("\n %d\t\t\t\t\t\t%s\n %d",i,pfirst->athlete[j].name,pfirst->athlete[j].score);break; } } } printf("\n\n\n\t\t\t\t\t\t按任意建 進入下一頁"); getchar(); pfirst = pfirst->next; } // clrscr(); printf("\n運動會結果:\n\n學校編號\t男運動員成績\t女運動員成績\t總分"); pfirst = head->next; while ( pfirst->next != NULL ) { printf("\n %d\t\t %d\t\t %d\t\t %d",pfirst->serial,pfirst->menscore,pfirst->womenscore,pfirst->totalscore); pfirst = pfirst->next; } printf("\n\n\n\t\t\t\t\t\t\t按任意建結束"); getchar(); } void create() { pfirst = (struct schoolstruct *)malloc(sizeof(struct schoolstruct)); pfirst->next = head->next ; head->next = pfirst ; pfirst->count = 1; pfirst->menscore = 0; pfirst->womenscore = 0; pfirst->totalscore = 0; } void Save() {FILE *fp; if((fp = fopen("school.dat","wb"))==NULL) {printf("can't open school.dat\n"); fclose(fp); return; } fwrite(pfirst,sizeof(SCH),10,fp); fclose(fp); printf("文件已經成功保存\n"); } void main() { system("cls"); printf("\n\t\t\t 運動會分數統計\n"); printf("輸入學校數目 (x>= 5):"); scanf("%d",&nsc); printf("輸入男選手的項目(x<=20):"); scanf("%d",&msp); printf("輸入女選手項目(<=20):"); scanf("%d",&wsp); ntsp = msp + wsp; phead = (int *)calloc(ntsp,sizeof(int)); pafirst = phead; pasecond = phead; input(); calculate(); output(); Save(); }
標簽: 源代碼
上傳時間: 2016-12-28
上傳用戶:150501