Computation of loudness (Zwicker model) according to ISO 532B / DIN 45631 norms. This model is valid for steady sounds. Code based on BASIC program published in the following article: Program for calculating loudness according to DIN 45 631 (ISO 532B)", E.Zwicker and H.Fastl, J.A.S.J (E) 12, 1 (1991).
上傳時(shí)間： 2016-11-14
上傳用戶：zztony16
void DFS(MGraph G, int i) { int j; visited[i] = TRUE; printf("%c ", G.vexs[i]); for (j=0; j<G.numVertexes; ++j) { if (G.arc[i][j]!=INFINITY && !visited[j]) DFS(G, j); } }
標(biāo)簽： 多項(xiàng)式 代碼 計(jì)算
上傳時(shí)間： 2016-12-28
上傳用戶：chenyameng
#include <malloc.h> #include <stdio.h> #include <stdlib.h> #include <string.h> #define NULL 0 #define MaxSize 30 typedef struct athletestruct /*運(yùn)動員*/ { char name[20]; int score; /*分?jǐn)?shù)*/ int range; /**/ int item; /*項(xiàng)目*/ }ATH; typedef struct schoolstruct /*學(xué)校*/ { int count; /*編號*/ int serial; /**/ int menscore; /*男選手分?jǐn)?shù)*/ int womenscore; /*女選手分?jǐn)?shù)*/ 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(yùn)動項(xiàng)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(gè)項(xiàng)目已經(jīng)存在請選擇其他的數(shù)字\n"); goto Is_Game_DoMain; } } } pafirst = pafirst + 1; if ( overgame > ntsp ) { printf("\n項(xiàng)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(xué)校數(shù)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(xù)輸入運(yùn)動項(xiàng)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(xué)校的結(jié)果成績:",pfirst->serial); printf("\n\n項(xiàng)目的數(shù)目\t學(xué)校的名字\t分?jǐn)?shù)"); 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(jìn)入下一頁"); getchar(); pfirst = pfirst->next; } // clrscr(); printf("\n運(yùn)動會結(jié)果:\n\n學(xué)校編號\t男運(yùn)動員成績\t女運(yùn)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(jié)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(jīng)成功保存\n"); } void main() { system("cls"); printf("\n\t\t\t 運(yùn)動會分?jǐn)?shù)統(tǒng)計(jì)\n"); printf("輸入學(xué)校數(shù)目 (x>= 5):"); scanf("%d",&nsc); printf("輸入男選手的項(xiàng)目(x<=20):"); scanf("%d",&msp); printf("輸入女選手項(xiàng)目(<=20):"); scanf("%d",&wsp); ntsp = msp + wsp; phead = (int *)calloc(ntsp,sizeof(int)); pafirst = phead; pasecond = phead; input(); calculate(); output(); Save(); }
標(biāo)簽： 源代碼
上傳時(shí)間： 2016-12-28
上傳用戶：150501
1．Describe a Θ(n lg n)-time algorithm that, given a set S of n integers and another integer x, determines whether or not there exist two elements in S whose sum is exactly x. (Implement exercise 2.3-7.) #include<stdio.h> #include<stdlib.h> void merge(int arr[],int low,int mid,int high){ int i,k; int *tmp=(int*)malloc((high-low+1)*sizeof(int)); int left_low=low; int left_high=mid; int right_low=mid+1; int right_high=high; for(k=0;left_low<=left_high&&right_low<=right_high;k++) { if(arr[left_low]<=arr[right_low]){ tmp[k]=arr[left_low++]; } else{ tmp[k]=arr[right_low++]; } } if(left_low<=left_high){ for(i=left_low;i<=left_high;i++){ tmp[k++]=arr[i]; } } if(right_low<=right_high){ for(i=right_low;i<=right_high;i++) tmp[k++]=arr[i]; } for(i=0;i<high-low+1;i++) arr[low+i]=tmp[i]; } void merge_sort(int a[],int p,int r){ int q; if(p<r){ q=(p+r)/2; merge_sort(a,p,q); merge_sort(a,q+1,r); merge(a,p,q,r); } } int main(){ int a[8]={3,5,8,6,4,1,1}; int i,j; int x=10; merge_sort(a,0,6); printf("after Merging-Sort:\n"); for(i=0;i<7;i++){ printf("%d",a[i]); } printf("\n"); i=0;j=6; do{ if(a[i]+a[j]==x){ printf("exist"); break; } if(a[i]+a[j]>x) j--; if(a[i]+a[j]<x) i++; }while(i<=j); if(i>j) printf("not exist"); system("pause"); return 0; }
上傳時(shí)間： 2017-04-01
上傳用戶：糖兒水嘻嘻
題目：古典問題：有一對兔子，從出生后第3個(gè)月起每個(gè)月都生一對兔子，小兔子長到第三個(gè)月后每個(gè)月又生一對兔子，假如兔子都不死，問每個(gè)月的兔子總數(shù)為多少？ //這是一個(gè)菲波拉契數(shù)列問題 public class lianxi01 { public static void main(String[] args) { System.out.println("第1個(gè)月的兔子對數(shù): 1"); System.out.println("第2個(gè)月的兔子對數(shù): 1"); int f1 = 1, f2 = 1, f, M=24; for(int i=3; i<=M; i++) { f = f2; f2 = f1 + f2; f1 = f; System.out.println("第" + i +"個(gè)月的兔子對數(shù): "+f2); } } } 【程序2】 題目：判斷101-200之間有多少個(gè)素?cái)?shù)，并輸出所有素?cái)?shù)。 程序分析：判斷素?cái)?shù)的方法：用一個(gè)數(shù)分別去除2到sqrt(這個(gè)數(shù))，如果能被整除， 則表明此數(shù)不是素?cái)?shù)，反之是素?cái)?shù)。 public class lianxi02 { public static void main(String[] args) { int count = 0; for(int i=101; i<200; i+=2) { boolean b = false; for(int j=2; j<=Math.sqrt(i); j++) { if(i % j == 0) { b = false; break; } else { b = true; } } if(b == true) {count ++;System.out.println(i );} } System.out.println( "素?cái)?shù)個(gè)數(shù)是: " + count); } } 【程序3】 題目：打印出所有的 "水仙花數(shù) "，所謂 "水仙花數(shù) "是指一個(gè)三位數(shù)，其各位數(shù)字立方和等于該數(shù)本身。例如：153是一個(gè) "水仙花數(shù) "，因?yàn)?53=1的三次方＋5的三次方＋3的三次方。 public class lianxi03 { public static void main(String[] args) { int b1, b2, b3;
上傳時(shí)間： 2017-12-24
上傳用戶：Ariza
#include "iostream" using namespace std; class Matrix { private: double** A; //矩陣A double *b; //向量b public: int size; Matrix(int ); ~Matrix(); friend double* Dooli(Matrix& ); void Input(); void Disp(); }; Matrix::Matrix(int x) { size=x; //為向量b分配空間并初始化為0 b=new double [x]; for(int j=0;j<x;j++) b[j]=0; //為向量A分配空間并初始化為0 A=new double* [x]; for(int i=0;i<x;i++) A[i]=new double [x]; for(int m=0;m<x;m++) for(int n=0;n<x;n++) A[m][n]=0; } Matrix::~Matrix() { cout<<"正在析構(gòu)中~~~~"<<endl; delete b; for(int i=0;i<size;i++) delete A[i]; delete A; } void Matrix::Disp() { for(int i=0;i<size;i++) { for(int j=0;j<size;j++) cout<<A[i][j]<<" "; cout<<endl; } } void Matrix::Input() { cout<<"請輸入A:"<<endl; for(int i=0;i<size;i++) for(int j=0;j<size;j++){ cout<<"第"<<i+1<<"行"<<"第"<<j+1<<"列:"<<endl; cin>>A[i][j]; } cout<<"請輸入b:"<<endl; for(int j=0;j<size;j++){ cout<<"第"<<j+1<<"個(gè):"<<endl; cin>>b[j]; } } double* Dooli(Matrix& A) { double *Xn=new double [A.size]; Matrix L(A.size),U(A.size); //分別求得U，L的第一行與第一列 for(int i=0;i<A.size;i++) U.A[0][i]=A.A[0][i]; for(int j=1;j<A.size;j++) L.A[j][0]=A.A[j][0]/U.A[0][0]; //分別求得U，L的第r行，第r列 double temp1=0,temp2=0; for(int r=1;r<A.size;r++){ //U for(int i=r;i<A.size;i++){ for(int k=0;k<r-1;k++) temp1=temp1+L.A[r][k]*U.A[k][i]; U.A[r][i]=A.A[r][i]-temp1; } //L for(int i=r+1;i<A.size;i++){ for(int k=0;k<r-1;k++) temp2=temp2+L.A[i][k]*U.A[k][r]; L.A[i][r]=(A.A[i][r]-temp2)/U.A[r][r]; } } cout<<"計(jì)算U得:"<<endl; U.Disp(); cout<<"計(jì)算L的:"<<endl; L.Disp(); double *Y=new double [A.size]; Y[0]=A.b[0]; for(int i=1;i<A.size;i++ ){ double temp3=0; for(int k=0;k<i-1;k++) temp3=temp3+L.A[i][k]*Y[k]; Y[i]=A.b[i]-temp3; } Xn[A.size-1]=Y[A.size-1]/U.A[A.size-1][A.size-1]; for(int i=A.size-1;i>=0;i--){ double temp4=0; for(int k=i+1;k<A.size;k++) temp4=temp4+U.A[i][k]*Xn[k]; Xn[i]=(Y[i]-temp4)/U.A[i][i]; } return Xn; } int main() { Matrix B(4); B.Input(); double *X; X=Dooli(B); cout<<"~~~~解得:"<<endl; for(int i=0;i<B.size;i++) cout<<"X["<<i<<"]:"<<X[i]<<" "; cout<<endl<<"呵呵呵呵呵"; return 0; }
標(biāo)簽： 道理特分解法
上傳時(shí)間： 2018-05-20
上傳用戶：Aa123456789
function [alpha,N,U]=youxianchafen2(r1,r2,up,under,num,deta) %[alpha,N,U]=youxianchafen2(a,r1,r2,up,under,num,deta) %該函數(shù)用有限差分法求解有兩種介質(zhì)的正方形區(qū)域的二維拉普拉斯方程的數(shù)值解 %函數(shù)返回迭代因子、迭代次數(shù)以及迭代完成后所求區(qū)域內(nèi)網(wǎng)格節(jié)點(diǎn)處的值 %a為正方形求解區(qū)域的邊長 %r1,r2分別表示兩種介質(zhì)的電導(dǎo)率 %up,under分別為上下邊界值 %num表示將區(qū)域每邊的網(wǎng)格剖分個(gè)數(shù) %deta為迭代過程中所允許的相對誤差限 n=num+1; %每邊節(jié)點(diǎn)數(shù) U(n,n)=0; %節(jié)點(diǎn)處數(shù)值矩陣 N=0; %迭代次數(shù)初值 alpha=2/(1+sin(pi/num));%超松弛迭代因子 k=r1/r2; %兩介質(zhì)電導(dǎo)率之比 U(1,1:n)=up; %求解區(qū)域上邊界第一類邊界條件 U(n,1:n)=under; %求解區(qū)域下邊界第一類邊界條件 U(2:num,1)=0;U(2:num,n)=0; for i=2:num U(i,2:num)=up-(up-under)/num*(i-1);%采用線性賦值對上下邊界之間的節(jié)點(diǎn)賦迭代初值 end G=1; while G>0 %迭代條件：不滿足相對誤差限要求的節(jié)點(diǎn)數(shù)目G不為零 Un=U; %完成第n次迭代后所有節(jié)點(diǎn)處的值 G=0; %每完成一次迭代將不滿足相對誤差限要求的節(jié)點(diǎn)數(shù)目歸零 for j=1:n for i=2:num U1=U(i,j); %第n次迭代時(shí)網(wǎng)格節(jié)點(diǎn)處的值 if j==1 %第n+1次迭代左邊界第二類邊界條件 U(i,j)=1/4*(2*U(i,j+1)+U(i-1,j)+U(i+1,j)); end if (j>1)&&(j U2=1/4*(U(i,j+1)+ U(i-1,j)+ U(i,j-1)+ U(i+1,j)); U(i,j)=U1+alpha*(U2-U1); %引入超松弛迭代因子后的網(wǎng)格節(jié)點(diǎn)處的值 end if i==n+1-j %第n+1次迭代兩介質(zhì)分界面（與網(wǎng)格對角線重合）第二類邊界條件 U(i,j)=1/4*(2/(1+k)*(U(i,j+1)+U(i+1,j))+2*k/(1+k)*(U(i-1,j)+U(i,j-1))); end if j==n %第n+1次迭代右邊界第二類邊界條件 U(i,n)=1/4*(2*U(i,j-1)+U(i-1,j)+U(i+1,j)); end end end N=N+1 %顯示迭代次數(shù) Un1=U; %完成第n+1次迭代后所有節(jié)點(diǎn)處的值 err=abs((Un1-Un)./Un1);%第n+1次迭代與第n次迭代所有節(jié)點(diǎn)值的相對誤差 err(1,1:n)=0; %上邊界節(jié)點(diǎn)相對誤差置零 err(n,1:n)=0; %下邊界節(jié)點(diǎn)相對誤差置零 G=sum(sum(err>deta))%顯示每次迭代后不滿足相對誤差限要求的節(jié)點(diǎn)數(shù)目G end
標(biāo)簽： 有限差分
上傳時(shí)間： 2018-07-13
上傳用戶：Kemin
反激式開關(guān)電源變壓器設(shè)計(jì)的詳細(xì)步驟85W反激變壓器設(shè)計(jì)的詳細(xì)步驟 1. 確定電源規(guī)格. 1).輸入電壓范圍Vin=90—265Vac; 2).輸出電壓/負(fù)載電流:Vout1=42V/2A, Pout=84W 3).轉(zhuǎn)換的效率=0.80 Pin=84/0.8=105W 2. 工作頻率，匝比, 最低輸入電壓和最大占空比確定. Ｖmos*0.8>Vinmax+n(Vo+Vf)600*0.8>373+n(42+1)得n<2.5Vd*0.8>Vinmax/n+Vo400*0.8>373/n+42得n>1.34 所以n取1.6最低輸入電壓Vinmin=√[(Vacmin√2)* (Vacmin√2)-2Pin(T/2-tc)/Cin=(90√2*90√2-2*105*(20/2-3)/0.00015=80V取:工作頻率fosc=60KHz, 最大占空比Dmax=n(Vo+Vf)/[n(Vo+Vf)+Vinmin]= 1.6(42+1)/[1.6(42+1)+80]=0.45 Ton(max)=1/f*Dmax=0.45/60000=7.5us 3. 變壓器初級峰值電流的計(jì)算. Iin-avg=1/3Pin/Vinmin=1/3*105/80=0.4AΔIp1=2Iin-avg/D=2*0.4/0.45=1.78AIpk1=Pout/?/Vinmin*D+ΔIp1=84/0.8/80/0.45=2.79A 4. 變壓器初級電感量的計(jì)算. 由式子Vdc=Lp*dip/dt,得: Lp= Vinmin*Ton(max)/ΔIp1 =80*0.0000075/1.78 =337uH 取Lp=337 uH 5.變壓器鐵芯的選擇. 根據(jù)式子Aw*Ae=Pt*1000000/[2*ko*kc*fosc*Bm*j*?],其中: Pt(標(biāo)稱輸出功率)= Pout=84W Ko(窗口的銅填充系數(shù))=0.4 Kc(磁芯填充系數(shù))=1(對于鐵氧體), 變壓器磁通密度Bm=1500Gs j(電流密度): j=4A/mm2;Aw*Ae=84*1000000/[2*0.4*1*60*103*1500Gs*4*0.80]=0.7cm4 考慮到繞線空間,選擇窗口面積大的磁芯,查表: ER40/45鐵氧體磁芯的有效截面積Ae=1.51cm2 ER40/45的功率容量乘積為 Ap = 3.7cm4 >0.7cm4 故選擇ER40/45鐵氧體磁芯. 6.變壓器初級匝數(shù) 1).由Np=Vinmin*Ton/[Ae*Bm],得: Np=80*7.5*10n-6/[1.52*10n-4*0.15] =26.31 取 Np =27T 7. 變壓器次級匝數(shù)的計(jì)算. Ns1(42v)=Np/n=27/1.6=16.875 取Ns1 = 17T Ns2(15v)=(15+1)* Ns1/(42+1)=6.3T 取Ns2 = 7T
標(biāo)簽： 開關(guān)電源 變壓器
上傳時(shí)間： 2022-04-15
上傳用戶：
本源代碼是基于STM32F4xx硬件平臺設(shè)計(jì)的貪吃蛇小游戲，主要難點(diǎn)在：隨機(jī)點(diǎn)產(chǎn)生、貪吃蛇轉(zhuǎn)向、貪吃蛇貪吃點(diǎn)；本部分主要接收產(chǎn)生隨機(jī)點(diǎn)，產(chǎn)生隨機(jī)點(diǎn)需要注意兩個(gè)方面：1、隨機(jī)點(diǎn)在有效的范圍內(nèi)；2、貪吃點(diǎn)與貪吃蛇不重合。產(chǎn)生隨機(jī)點(diǎn)主要有兩個(gè)函數(shù)，分別如下：//隨機(jī)數(shù)產(chǎn)生任務(wù)void rng_chansheng(void *p_arg){OS_ERR err;while(1){OSSemPend(&RNG_SEM,0,OS_OPT_PEND_BLOCKING,0,&err);zou.x = RNG_Get_RandomRange(0,50)*8 + 40;zou.y = RNG_Get_RandomRange(0,50)*8 + 260;lcd_fangkuan(zou.x,zou.y,zou.x+SHE_FAANGKUAN_SIZE,zou.y+SHE_FAANGKUAN_SIZE);OSTimeDlyHMSM(0,0,0,500,OS_OPT_TIME_HMSM_STRICT,&err); //延時(shí)500ms}}//往下方向畫一個(gè)實(shí)心的正方形，代表貪食蛇的一段void lcd_fangkuan(u16 x1,u16 y1,u16 x2 ,u16 y2){u16 i,j;u16 xx,yy;if(((x2 - x1) != SHE_FAANGKUAN_SIZE)||((y2 - y1) != SHE_FAANGKUAN_SIZE))return ;if(x1 > x2) {xx = x1;x1 = x2;x2 = xx;}if(y1 > y2){yy = y1;y1 = y2;y2 = yy;}if((y1 < 260)|| (y2 > 660)||(x1 < 40)||(x2 > 448)){game_yun_error = 1;LCD_ShowString(150,300,500,24,24,"GAME OVER!!");return ;}for(i=x1; i<x2; i++){for(j=y1; j<y2; j++){LCD_DrawPoint(i,j);}}}
上傳時(shí)間： 2022-08-10
上傳用戶：
圖書管理系統(tǒng)畢業(yè)設(shè)計(jì)+源碼.rar j++ 與 access數(shù)據(jù)庫 供學(xué)習(xí)使用
上傳時(shí)間： 2016-01-10
上傳用戶：stampede
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