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資料型態 陣列名稱 [ 長度 ]; 資料型態 陣列名稱 [ 長度 ]={ 值 1, 值 2,….}; PowerPoint PPT Presentation

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陣列

• 資料型態 陣列名稱[長度];

• 資料型態 陣列名稱[長度]={值1,值2,….};

int a[3];

a[0]=2;

a[1]=1;

a[2]=5;

int a[3]={2,1,5};

//排序;由小排到大

# include <iostream.h>

void main()

{

int i,j,b;

int a[5]={2,6,3,1,5};

for (i=0;i<(5-1);i++)

{

for(j=i+1;j<5;j++)

{

if(a[i]>a[j]) {b=a[i];a[i]=a[j];a[j]=b;}

}

}

for(i=0;i<5;i++)cout<<a[i];

cout<<“\n”;

}

# include <iostream.h>

void main()

{

int i,sum=0;

int a[5]={2,6,3,1,5};

double p;

for (i=0;i<(5-1);i++)sum=sum+a[i];

for(i=0;i<4;i++)

{

p=double(a[i])/sum;

cout<<a[i]<<“百分比”<<p<<“\n”;

}

}

字串

• char letter;

• char string[4];

letter=‘a’;

string=“C++”;//錯誤寫法

char string[4]=“C++”;//字串陣列的長度要大 於字串至少1

char x[4];

cin>>x;

字串

char x[4]=“C++”; //字串陣列的長度要大 於字串至少1

X[0] x[1] x[2] x[3]

//’\0’字串結束字元 …’0’是零

輸入一字串,將此字串以反向輸出。例如:輸入steven輸出nevets

#include<iostream.h>

void main()

{

char x[50];

int i,L=0;

cin>>x;

for(i=0;i<50;i++)

{

if(x[i]==‘\0’)break;

L=L+1;

}

for(i=L-1;i>=0;i--)cout<<x[i];

}

輸入身份證字號

//判斷身分字號的長度

#include<iostream.h>

void main()

{

char x[20];

int i,L,right;

L=0;right=0;

while(right!=1){

cin >>x;

for(i=0;i<20;i++){

if (x[i]=='\0')break;

cout<<x[i];

L=L+1;}

if(L==10))right=1;

else cout<<“長度輸入錯誤\n”;}

}

輸入身份證字號

//判斷輸入的身份證字號第一個是否為英文字母

#include<iostream.h>

void main()

{

char x[20];

int right;

right=0;

while(right!=1)

{

cin >>x;

if((x[0]<='z'&& x[0]>='a') || (x[0]<='Z' && x[0]>='A'))right=1;

else cout<<“輸入錯誤\n”;

}

}

輸入身份證字號

//判斷輸入的身份證字號的阿拉伯數

#include<iostream.h>

void main()

{

char x[20];

int right=0,L=10;

while(right!=1)

{

cin >>x;

for(i=1;i<L;i++)

{

if(x[i]<=‘9’&& x[i]>=‘0’)right=1;

else {right=0;cout<<“輸入錯誤\n”;break;}

}

}

}

輸入身份證字號

//判斷輸入的身份證字號的阿拉伯數

#include<iostream.h>

void main()

{

char x[20];

int right=0,L,i;

while(right!=1){

L=0;

cin >>x;

for(i=0;i<20;i++){if (x[i]=='\0')break;cout<<x[i];L=L+1;}

if(L==10)right=1;

if(right==1){right=0; if((x[0]<='z'&& x[0]>='a') || (x[0]<='Z' && x[0]>='A'))right=1;}

if(right==1){right=0; for(i=1;i<L;i++)if(x[i]<=‘9’&& x[i]>=‘0’)right=1; else{right=-1;break;}}

if(right!=1)cout<<“輸入錯誤\n”;}

}

‘A’,’B’,…..’Z’→10,11,……33

W的位置在Y的後面

A→10拆成x1=1;x2=0

A1=1;A2=2……A9=9;

Y=X1+9*X2+8*A1+7*A2+….+2*A7+A8+A9

Y能被10整除 RIGHT

ASCII 0(零)→48

A→65 a→97

ASCII 0(零)→48

A→65 a→97

if(right==1){right=0;

if(x[0]>=97)z=x[0]-87;else x[0]=z-55;

z1=z/10;

z2=z%10;

y=z1+z2*9;

for(i=1;i<(L-1);i++){y=y+(x[i]-48)*(L-1-i);}

y=y+x[L-1]-48;

if(y%10==0)right=1;}

# include <iostream.h>

void main()

{

int i,j,k,n;

int d[12]={31,28,31,30,31,30,31,31,30,31,30,31};

char m[12][7]={"一月","二月","三月","四月","五月","六月","七月","八月“,"九月“,"十月","十一月","十二月"};

k=1;//2007年一月一日星期 一

for(i=0;i<12;i++)

{

cout<<m[i]<<"\n";

cout<<" 日 一 二 三 四 五 六\n";

for(n=1;n<=k;n++) cout<<" ";

for(j=1;j<=d[i];j++)

{

if (j<10)cout<<" "<<j;

else cout<<" "<<j;

k++;

k=k%7;

if(k==0) cout<<"\n";

}

cout<<"\n\n";

}

}

函數的定義int checkLen(char x[20]){int Len;return Len;}void main(){L=checkLen(x);}

int checkLen(char x[20]){int i,L=0;for(i=0;i<20;i++){if (x[i]=='\0') break; cout<<x[i];L=L+1;}if(L==10)right=1;return right;}int checkFir(char x[20],int right){if((x[0]<='z'&& x[0]>='a') || (x[0]<='Z' && x[0]>='A'))right=1;return right;}void main(){.right=checkLen(x,right);if(right==1){right=0;right=checkFir(x,right);}if(right==1){right=0; right=checkNum(x,right);}if(right==1){right=0; right=checkID(x,right); } }

Taylor’s series

Euler’s Method

檔案的輸出(寫入檔案)#include<fstream.h>ofstream fop(“test.dat”);// 用ofstream宣告一物件fopfop<<x<<‘ ‘<<y<<‘\n’; fop.close();

#include<iostream.h>

#include<fstream.h>

void main()

{

double x0,y0,y1,f,h=0.01;

x0=0;

y0=1;

ofstream fop("test.txt");

while(x0<=2.0)

{

fop<<x0<<' '<<y0<<'\n';

f=-2*x0*y0;

y1=y0+h*f;

x0=x0+h;

y0=y1;

}

fop.close();

}

y1=y

y2=y1’

y3=y2’=-2*y2-90*y1

Euler’s Method

#include<iostream.h>

#include<fstream.h>

void main()

{

double x0,y10,y20,y30,h=0.01,y11,y21;

x0=0;

y10=0.15;

y20=0;

ofstream fop("test.txt");

while(x0<=5.0)

{

y30=-2*y20-90*y10;

fop<<x0<<' '<<y10<<'\n';

y11=y10+h*y20;

y21=y20+h*y30;

x0=x0+h;

y10=y11;

y20=y21;

}

fop.close();

}