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  • API函數(shù)詳解winAPI32

    API函數(shù)詳解winAPI32.chm 表1 SetWindowPos函數(shù)的參數(shù)解釋 參數(shù)名 參數(shù)含義 hwnd 要移動的窗口的句柄(可以用窗體的hwnd屬性) hWndInsertAfter 關于如何在Z軸上放置窗口的標記(具體見表2) x 相當于窗口的Left屬性 y 相當于窗口的Top屬性 cx 相當于窗口的Right屬性 cy 相當于窗口的Bottom屬性 wFlags 關于如何移動窗口的標記(具體見表3)

    標簽: winAPI API 32 函數(shù)

    上傳時間: 2019-12-01

    上傳用戶:gdqiutao

  • matlab神經(jīng)網(wǎng)絡算法通信信號調(diào)制識別

    %========================開始提取加噪信號的各類特征值================================ for n=1:1:50;     m=n*Ns;     x=(n-1)*Ns; for i=x+1:m;          %提取加噪信號'signal_with_noise=y+noise'的前256個元素,抽取50次     y0(i)=signal_with_noise(i); end Y=fft(y0);        %對調(diào)制信號進行快速傅里葉算法(離散) y1=hilbert(y0) ;    %調(diào)制信號實部的解析式  factor=0;        %開始求零中心歸一化瞬時幅度譜密度的最大值gamma_max for i=x+1:m;    factor=factor+y0(i); end ms=factor/(m-x); an_i=y0./ms; acn_i=an_i-1; end gamma_max=max(fft(acn_i.*acn_i))/Ns

    標簽: matlab 神經(jīng)網(wǎng)絡算法 通信信號 調(diào)制識別

    上傳時間: 2020-04-07

    上傳用戶:如拷貝般復制

  • VHDL4選1數(shù)據(jù)選擇器

    VHDL編寫的4選一數(shù)據(jù)選擇器 entity mux41a is        port(a,b:in std_logic;                s1,s2,s3,s4:in std_logic;                y: out std_logic); end entity mux41a; architecture one of mux41a is signal ab:std_logic_vector(1 downto 0);

    標簽: VHDL 數(shù)據(jù)選擇器

    上傳時間: 2020-05-15

    上傳用戶:cdga

  • ComSoc+Guide+to+Next+Generation+Optical+Transport

    Many times I have been asked to explain “ briefl y ” how SDH, SONET, and the OTN “ exactly ” work. The questions came mainly from new colleagues, stu- dents, and users of these technologies, personally or via the usenet newsgroup comp.dcom.sdh - sonet. I could have referred them to the standards documents, but to provide a more consistent and clear answer I decided to write this pocket guide. The objective of this book is that it can be used both as an introduction as well as a reference guide to these technologies and their spe- cifi c standards documents.

    標簽: Generation Transport Optical ComSoc Guide

    上傳時間: 2020-05-27

    上傳用戶:shancjb

  • lagr.m

    function y=lagr(x0,y0,x) %x0,y0為節(jié)點 %x是插值點 n=length(x0); m=length(x); for i=1:m z=x(i); s=0.0; for k=1:n p=1.0; for j=1:n if j~=k p=p*(z-x0(j))/(x0(k)-x0(j)); end end s=p*y0(k)+s; end y(i)=s; end

    標簽: lagr

    上傳時間: 2020-06-09

    上傳用戶:shiyc2020

  • Digital+Control+Applications

    My association with the theory of controls in continuous time started during my studies at the Indian Institute of Technology, Kharagpur, India, in 1974 as an undergraduate student in the Controls and Power program. The initial introduction by Professors Kesavamurthy, Y. P. Singh, and Rajagopalan laid the foundation for a good basic understanding of the subject matter. This pursuit and further advanced study in the field of digital controls continued during my days as a graduate student in the Electrical and Systems Engineering Department at the University of Connecticut in Storrs, from 1983 to 1988.

    標簽: Applications Digital Control

    上傳時間: 2020-06-10

    上傳用戶:shancjb

  • 二叉樹子系統(tǒng)

    #include<stdio.h> #define TREEMAX 100 typedef struct  BT { char data; BT *lchild; BT *rchild; }BT; BT *CreateTree(); void Preorder(BT *T); void Postorder(BT *T); void Inorder(BT *T); void Leafnum(BT *T); void Nodenum(BT *T); int TreeDepth(BT *T); int count=0; void main() { BT *T=NULL; char ch1,ch2,a; ch1='y'; while(ch1=='y'||ch1=='y') { printf("\n"); printf("\n\t\t             二叉樹子系統(tǒng)"); printf("\n\t\t*****************************************"); printf("\n\t\t           1---------建二叉樹            "); printf("\n\t\t           2---------先序遍歷            "); printf("\n\t\t           3---------中序遍歷            "); printf("\n\t\t           4---------后序遍歷            "); printf("\n\t\t           5---------求葉子數(shù)            "); printf("\n\t\t           6---------求結(jié)點數(shù)            "); printf("\n\t\t           7---------求樹深度            "); printf("\n\t\t           0---------返    回            "); printf("\n\t\t*****************************************"); printf("\n\t\t      請選擇菜單號 (0--7)"); scanf("%c",&ch2); getchar(); printf("\n"); switch(ch2) { case'1': printf("\n\t\t請按先序序列輸入二叉樹的結(jié)點:\n"); printf("\n\t\t說明:輸入結(jié)點(‘0’代表后繼結(jié)點為空)后按回車。\n"); printf("\n\t\t請輸入根結(jié)點:"); T=CreateTree(); printf("\n\t\t二叉樹成功建立!\n");break; case'2': printf("\n\t\t該二叉樹的先序遍歷序列為:"); Preorder(T);break; case'3': printf("\n\t\t該二叉樹的中序遍歷序列為:"); Inorder(T);break; case'4': printf("\n\t\t該二叉樹的后序遍歷序列為:"); Postorder(T);break; case'5': count=0;Leafnum(T); printf("\n\t\t該二叉樹有%d個葉子。\n",count);break; case'6': count=0;Nodenum(T); printf("\n\t\t該二叉樹總共有%d個結(jié)點。\n",count);break; case'7': printf("\n\t\t該樹的深度為:%d",TreeDepth(T)); break; case'0': ch1='n';break; default: printf("\n\t\t***請注意:輸入有誤!***"); } if(ch2!='0') { printf("\n\n\t\t按【Enter】鍵繼續(xù),按任意鍵返回主菜單!\n"); a=getchar(); if(a!='\xA') { getchar(); ch1='n'; } } } } BT *CreateTree() { BT *t; char x; scanf("%c",&x); getchar(); if(x=='0') t=NULL; else { t=new BT; t->data=x; printf("\n\t\t請輸入%c結(jié)點的左子結(jié)點:",t->data);         t->lchild=CreateTree(); printf("\n\t\t請輸入%c結(jié)點的右子結(jié)點:",t->data);         t->rchild=CreateTree();     } return t; } void Preorder(BT *T) { if(T) { printf("%3c",T->data); Preorder(T->lchild); Preorder(T->rchild); } } void Inorder(BT *T) { if(T) { Inorder(T->lchild); printf("%3c",T->data); Inorder(T->rchild); } } void Postorder(BT *T) { if(T) { Postorder(T->lchild); Postorder(T->rchild); printf("%3c",T->data); } } void Leafnum(BT *T) { if(T) { if(T->lchild==NULL&&T->rchild==NULL) count++; Leafnum(T->lchild); Leafnum(T->rchild); } } void Nodenum(BT *T) { if(T) { count++; Nodenum(T->lchild); Nodenum(T->rchild); } } int TreeDepth(BT *T) { int ldep,rdep; if(T==NULL) return 0; else { ldep=TreeDepth(T->lchild); rdep=TreeDepth(T->rchild); if(ldep>rdep) return ldep+1; else return rdep+1; } }

    標簽: 二叉樹 子系統(tǒng)

    上傳時間: 2020-06-11

    上傳用戶:ccccy

  • 數(shù)組子系統(tǒng)

    #include <stdio.h> #include <stdlib.h> #define SMAX 100 typedef struct SPNode { int i,j,v; }SPNode; struct sparmatrix { int rows,cols,terms; SPNode data [SMAX]; }; sparmatrix CreateSparmatrix() { sparmatrix A; printf("\n\t\t請輸入稀疏矩陣的行數(shù),列數(shù)和非零元素個數(shù)(用逗號隔開):"); scanf("%d,%d,%d",&A.cols,&A.terms); for(int n=0;n<=A.terms-1;n++) { printf("\n\t\t輸入非零元素值(格式:行號,列號,值):"); scanf("%d,%d,%d",&A.data[n].i,&A.data[n].j,&A.data[n].v); } return A; } void ShowSparmatrix(sparmatrix A) { int k; printf("\n\t\t"); for(int x=0;x<=A.rows-1;x++) { for(int y=0;y<=A.cols-1;y++) { k=0; for(int n=0;n<=A.terms-1;n++) { if((A.data[n].i-1==x)&&(A.data[n].j-1==y)) { printf("%8d",A.data[n].v); k=1; } } if(k==0) printf("%8d",k); } printf("\n\t\t"); } } void sumsparmatrix(sparmatrix A) { SPNode *p; p=(SPNode*)malloc(sizeof(SPNode)); p->v=0; int k; k=0; printf("\n\t\t"); for(int x=0;x<=A.rows-1;x++) { for(int y=0;y<=A.cols-1;y++) { for(int n=0;n<=A.terms;n++) { if((A.data[n].i==x)&&(A.data[n].j==y)&&(x==y)) { p->v=p->v+A.data[n].v; k=1; } } } printf("\n\t\t"); } if(k==1) printf("\n\t\t對角線元素的和::%d\n",p->v); else printf("\n\t\t對角線元素的和為::0"); } int main() { int ch=1,choice; struct sparmatrix A; A.terms=0; while(ch) { printf("\n"); printf("\n\t\t      稀疏矩陣的三元組系統(tǒng)       "); printf("\n\t\t*********************************"); printf("\n\t\t      1------------創(chuàng)建          "); printf("\n\t\t      2------------顯示          "); printf("\n\t\t      3------------求對角線元素和"); printf("\n\t\t      4------------返回          "); printf("\n\t\t*********************************"); printf("\n\t\t請選擇菜單號(0-3):"); scanf("%d",&choice); switch(choice) { case 1: A=CreateSparmatrix(); break; case 2: ShowSparmatrix(A); break; case 3: SumSparmatrix(A); break; default: system("cls"); printf("\n\t\t輸入錯誤!請重新輸入!\n"); break; } if (choice==1||choice==2||choice==3) { printf("\n\t\t"); system("pause"); system("cls"); } else system("cls"); } }

    標簽: 數(shù)組 子系統(tǒng)

    上傳時間: 2020-06-11

    上傳用戶:ccccy

  • 某疾病C代碼

    某疾病發(fā)生率y‰和年齡段x(每五年為一段,例如0~5歲為第一段..)

    標簽: C代碼

    上傳時間: 2020-06-14

    上傳用戶:ll12346

  • 計算方法上機實習題目

    第一種邊界條件下的三次樣條插值問題(高斯消元法) 題目 計算方法上機實習題目(二) ——第一種邊界條件下的三次樣條插值問題(高斯消元法) 已知直升飛機旋轉(zhuǎn)機翼外形曲線輪廓線上的某些型值點(見表)及端點處的一階導數(shù)值 y' (x ) = 1.86548, y' (x ) = -0.046115

    標簽: 計算方法上機實習題目

    上傳時間: 2020-06-27

    上傳用戶:zg250

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