求標準偏差 > function c=myfunction(x) > [m,n]=size(x) > t=0 > for i=1:numel(x) > t=t+x(i)*x(i) > end > c=sqrt(t/(m*n-1)) function c=myfunction(x) [m,n]=size(x) t=0 for i=1:m for j=1:n t=t+x(i,j)*x(i,j) end end c=sqrt(t/(m*n-1
標簽: gt myfunction function numel
上傳時間: 2014-01-15
上傳用戶:hongmo
求標準偏差 > function c=myfunction(x) > [m,n]=size(x) > t=0 > for i=1:numel(x) > t=t+x(i)*x(i) > end > c=sqrt(t/(m*n-1)) function c=myfunction(x) [m,n]=size(x) t=0 for i=1:m for j=1:n t=t+x(i,j)*x(i,j) end end c=sqrt(t/(m*n-1
標簽: gt myfunction function numel
上傳時間: 2013-12-26
上傳用戶:dreamboy36
求標準偏差 > function c=myfunction(x) > [m,n]=size(x) > t=0 > for i=1:numel(x) > t=t+x(i)*x(i) > end > c=sqrt(t/(m*n-1)) function c=myfunction(x) [m,n]=size(x) t=0 for i=1:m for j=1:n t=t+x(i,j)*x(i,j) end end c=sqrt(t/(m*n-1
標簽: gt myfunction function numel
上傳時間: 2016-06-28
上傳用戶:change0329
求標準偏差 > function c=myfunction(x) > [m,n]=size(x) > t=0 > for i=1:numel(x) > t=t+x(i)*x(i) > end > c=sqrt(t/(m*n-1)) function c=myfunction(x) [m,n]=size(x) t=0 for i=1:m for j=1:n t=t+x(i,j)*x(i,j) end end c=sqrt(t/(m*n-1
標簽: gt myfunction function numel
上傳時間: 2014-09-03
上傳用戶:jjj0202
設有由n個不相同的整數組成的數列,記為: a(1)、a(2)、……、a(n)且a(i)<>a(j) (i<>j) 例如3,18,7,14,10,12,23,41,16,24。 若存在i1<i2<i3< … < ie 且有a(i1)<a(i2)< … <a(ie)則稱為長度為e的不下降序列。如上例中3,18,23,24就是一個長度為4的不下降序列,同時也有3,7,10,12,16,24長度為6的不下降序列。程序要求,當原數列給出之后,求出最長的不下降序列。
上傳時間: 2013-12-14
上傳用戶:tonyshao
Hopfield 網——擅長于聯想記憶與解迷路 實現H網聯想記憶的關鍵,是使被記憶的模式樣本對應網絡能量函數的極小值。 設有M個N維記憶模式,通過對網絡N個神經元之間連接權 wij 和N個輸出閾值θj的設計,使得: 這M個記憶模式所對應的網絡狀態正好是網絡能量函數的M個極小值。 比較困難,目前還沒有一個適應任意形式的記憶模式的有效、通用的設計方法。 H網的算法 1)學習模式——決定權重 想要記憶的模式,用-1和1的2值表示 模式:-1,-1,1,-1,1,1,... 一般表示: 則任意兩個神經元j、i間的權重: wij=∑ap(i)ap(j),p=1…p; P:模式的總數 ap(s):第p個模式的第s個要素(-1或1) wij:第j個神經元與第i個神經元間的權重 i = j時,wij=0,即各神經元的輸出不直接返回自身。 2)想起模式: 神經元輸出值的初始化 想起時,一般是未知的輸入。設xi(0)為未知模式的第i個要素(-1或1) 將xi(0)作為相對應的神經元的初始值,其中,0意味t=0。 反復部分:對各神經元,計算: xi (t+1) = f (∑wijxj(t)-θi), j=1…n, j≠i n—神經元總數 f()--Sgn() θi—神經元i發火閾值 反復進行,直到各個神經元的輸出不再變化。
上傳時間: 2015-03-16
上傳用戶:JasonC
#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
#include "string.h" #include "ctype.h" #include "stdio.h" search(char pd[]) {FILE *fp; int time=0,i=0,j=0,add[80],k=0,m; char *ch, str[900]; m=strlen(pd); if((fp=fopen("haha.txt","r"))==NULL) { printf("Cannot open this file\n"); exit(0); } for(;!feof(fp);i++) { str[i]=fgetc(fp); if(tolower(str[i])==tolower(pd[k])) {k++; if(k==m) if(!isalpha(i-m)&&!isalpha((str[i++]=fgetc(fp)))) { time++; add[j]=i-m+1; j++; k=0; } else k=0; } } if(time) { printf("The time is:%d\n",time); printf("The adders is:\n"); for(i=0;i<j;i++) printf("%5d",add[i]); if(i%5==0) printf("\n"); getch(); fclose(fp); } else printf("Sorry!Cannot find the word(^_^)"); } main() { char pd[10],choose='y'; int flag=1; while(flag) {printf("In put the word you want to seqarch:"); scanf("%s",pd); search(strlwr(pd)); printf("\nWould you want to continue?(Y/N):"); getchar(); scanf("%c",&choose); if((tolower(choose))=='n') flag=0; else flag=1; } printf("Thanks for your using!Bye-bye!\n"); getch(); }
標簽: 學生專用
上傳時間: 2016-12-29
上傳用戶:767483511
# include<stdio.h> # include<math.h> # define N 3 main(){ float NF2(float *x,float *y); float A[N][N]={{10,-1,-2},{-1,10,-2},{-1,-1,5}}; float b[N]={7.2,8.3,4.2},sum=0; float x[N]= {0,0,0},y[N]={0},x0[N]={}; int i,j,n=0; for(i=0;i<N;i++) { x[i]=x0[i]; } for(n=0;;n++){ //計算下一個值 for(i=0;i<N;i++){ sum=0; for(j=0;j<N;j++){ if(j!=i){ sum=sum+A[i][j]*x[j]; } } y[i]=(1/A[i][i])*(b[i]-sum); //sum=0; } //判斷誤差大小 if(NF2(x,y)>0.01){ for(i=0;i<N;i++){ x[i]=y[i]; } } else break; } printf("經過%d次雅可比迭代解出方程組的解:\n",n+1); for(i=0;i<N;i++){ printf("%f ",y[i]); } } //求兩個向量差的二范數函數 float NF2(float *x,float *y){ int i; float z,sum1=0; for(i=0;i<N;i++){ sum1=sum1+pow(y[i]-x[i],2); } z=sqrt(sum1); return z; }
上傳時間: 2019-10-13
上傳用戶:大萌萌撒
function y=lagr(x0,y0,x) %x0,y0為節點 %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