C++編程實(shí)例參考模板

1、例313求Fibonacci數(shù)列前40個(gè)數(shù)。這個(gè)數(shù)列有如下特點(diǎn)：第1、2個(gè)數(shù)為1、1。從第3個(gè)數(shù)開始，每個(gè)數(shù)是其前面兩個(gè)數(shù)之和。#include <iostream>#include <iomanip>using namespace std;int main( )long f1,f2;int i;f1=f2=1;for( i=1; i<=20; i+ )/設(shè)備輸出字段寬度為12，每次輸出兩個(gè)數(shù)cout<<setw(12)<<f1<<setw(12)<<f2; /每輸出完4個(gè)數(shù)后換行，使每行輸出4個(gè)數(shù)if(i%2=0)

2、 cout<<endl;f1=f1+f2; /左邊的f1代表第3個(gè)數(shù)，是第1、2個(gè)數(shù)之和f2=f2+f1; /左邊的f2代表第4個(gè)數(shù)，是第2、3個(gè)數(shù)之和return 0;例314100200間的素?cái)?shù)#include <iostream>1 / 56#include <cmath>#include <iomanip>using namespace std;int main() int m,k,i,n=0; bool prime; for(m=101;m<=200;m=m+2) prime=true;k=int(sqrt(m); for(i=2

3、;i<=k;i+)if(m%i=0) prime =false; break; if(prime) cout <<setw(5) <<m; n=n+1; if(n%10=0) cout <<endl; cout <<endl;return 0;例315譯密碼#include <iostream>using namespace std;int main()char c; while (c=getchar()!='n')if(c>='a'&&c<='z')|

4、 (c>='A'&&c<='Z') c=c+4; if(c>'Z'&&c<='Z'+4|c>'z')c=c-26;cout<<c;cout<<endl;return 0;習(xí)題：16、統(tǒng)計(jì)個(gè)數(shù)#include <iostream>using namespace std;int main ()char c; int letters=0,space=0,digit=0,other=0; cout<<"e

5、nter one line:"<<endl; while(c=getchar()!='n') if (c>='a' && c<='z'|c>='A' && c<='Z') letters+; else if (c=' ') space+; else if (c>='0' && c<='9') digit+; else other+; cout<<&qu

6、ot;letter:"<<letters<<", space:"<<space<<", digit:"<<digit<<", other:"<<other<<endl; return 0; 17、求Sn=a+aa+aaa+aaa（n個(gè)a）的值，n由鍵盤輸入。#include<iostream>using namespace std;int main ()int a,n,i=1,sn=0,tn=0; cout<&l

7、t;"a,n=:" cin>>a>>n; while (i<=n) tn=tn+a; /賦值后的tn為i個(gè)a組成數(shù)的值 sn=sn+tn; /賦值后的sn為多項(xiàng)式前i項(xiàng)之和 a=a*10; +i; cout<<"a+aa+aaa+.="<<sn<<endl; return 0; 18、.求1！+2！+20！、#include <iostream>using namespace std;int main ()float s=0,t=1; int n; for (n=1;n<

8、=20;n+) t=t*n; / 求n! s=s+t; / 將各項(xiàng)累加 cout<<"1!+2!+.+20!="<<s<<endl; return 0; 19、求“水仙花數(shù)”例153=13+33+53#include <iostream>using namespace std;int main ()int i,j,k,n; cout<<"narcissus numbers are:"<<endl; for (n=100;n<1000;n+) i=n/100; j=n/10-i*

9、10; k=n%10; if (n = i*i*i + j*j*j + k*k*k) cout<<n<<" " cout<<endl;return 0; 20、找出1000之內(nèi)所有的完數(shù)例如6的因子是1.2.3而6=1+2+3，所以6是完數(shù)#include <iostream>using namespace std; int main() int m,s,i; for (m=2;m<1000;m+) s=0; for (i=1;i<m;i+) if (m%i)=0) s=s+i; if(s=m) cout<&

10、lt;m<<" is a完數(shù)"<<endl; cout<<"its factors are:" for (i=1;i<m;i+) if (m%i=0) cout<<i<<" " cout<<endl; return 0; 21、2/1, 3/2, 5/3, 8/5, 13/8, 21/13,前20項(xiàng)之和#include <iostream>using namespace std;int main() int i,t,n=20; double a=

11、2,b=1,s=0; for (i=1;i<=n;i+) s=s+a/b; t=a; a=a+b; / 將前一項(xiàng)分子與分母之和作為下一項(xiàng)的分子 b=t; / 將前一項(xiàng)的分子作為下一項(xiàng)的分母 cout<<"sum="<<s<<endl; return 0; 24、輸出圖形#include <iostream>using namespace std;int main() int i,k; for (i=0;i<=3;i+) / 輸出上面4行*號(hào) for (k=0;k<=2*i;k+) cout<<&q

12、uot;*" / 輸出*號(hào) cout<<endl; /輸出完一行*號(hào)后換行 for (i=0;i<=2;i+) / 輸出下面3行*號(hào) for (k=0;k<=4-2*i;k+) cout<<"*" / 輸出*號(hào) cout<<endl; / 輸出完一行*號(hào)后換行 return 0;習(xí)題：1、求最大公約數(shù)和最小公倍數(shù)，用主函數(shù)條用兩個(gè)函數(shù) #include <iostream>using namespace std;int main() int hcf(int,int); int lcd(int,int,in

13、t); int u,v,h,l; cin>>u>>v; h=hcf(u,v); cout<<"H.C.F="<<h<<endl; l=lcd(u,v,h); cout<<"L.C.D="<<l<<endl; return 0; int hcf(int u,int v) int t,r; if (v>u) t=u;u=v;v=t; while (r=u%v)!=0) u=v; v=r; return(v); int lcd(int u,int v,int

14、h) return(u*v/h); 3、判別素?cái)?shù)的函數(shù) #include <iostream>using namespace std;int main() int prime(int); /* 函數(shù)原型聲明 */ int n; cout<<"input an integer:" cin>>n; if (prime(n) cout<<n<<" is a prime."<<endl; else cout<<n<<" is not a prime.&quo

15、t;<<endl; return 0; int prime(int n) int flag=1,i; for (i=2;i<n/2 && flag=1;i+) if (n%i=0) flag=0; return(flag); 4、.求a！+b！+c！的值，用一個(gè)函數(shù)fac（n）求n！。4#include <iostream>using namespace std;int main() int fac(int); int a,b,c,sum=0; cout<<"enter a,b,c:" cin>>a>

16、;>b>>c; sum=sum+fac(a)+fac(b)+fac(c); cout<<a<<"!+"<<b<<"!+"<<c<<"!="<<sum<<endl; return 0; int fac(int n) int f=1; for (int i=1;i<=n;i+) f=f*i; return f; 7、.驗(yàn)證哥德巴赫猜想：一個(gè)不小于6的偶數(shù)可以表示為兩個(gè)素?cái)?shù)之和。7#include <iostream

17、>#include <cmath>using namespace std;int main()void godbaha(int); int n; cout<<"input n:" cin>>n; godbaha(n); return 0;void godbaha(int n) int prime(int); int a,b; for(a=3;a<=n/2;a=a+2) if(prime(a) b=n-a; if (prime(b) cout<<n<<"="<<a<&

18、lt;"+"<<b<<endl; int prime(int m)int i,k=sqrt(m); for(i=2;i<=k;i+)if(m%i=0) break; if (i>k) return 1; else return 0;11、用遞歸方法求f（n）=12+22+n2#include <iostream>using namespace std;int main() int f(int); int n,s; cout<<"input the number n:" cin>>n;

19、 s=f(n); cout<<"The result is "<<s<<endl; return 0; int f(int n) if (n=1) return 1; else return (n*n+f(n-1);例52 用數(shù)組Fibonacci數(shù)列#include <iostream> #include <iomanip>using namespace std;int main () int i; int f20=1,1; for(i=2;i<20;i+) fi=fi-2+fi-1; for(i=0;i&

20、lt;20;i+) if(i%5=0)cout<<endl; cout<<setw(8)<<fi; cout<<endl; return 0;例5.3用冒泡法對(duì)10個(gè)數(shù)排序#include <iostream>using namespace std;int main( ) int a10; int i,j,t; cout<<"input 10 numbers ："<<endl; for (i=0;i<10;i+) /輸入a0a9cin>>ai; cout<<en

21、dl; for (j=0;j<9;j+) /共進(jìn)行9趟比較 for(i=0;i<9-j;i+) /在每趟中要進(jìn)行(10-j)次兩兩比較 if (ai>ai+1) /如果前面的數(shù)大于后面的數(shù) t=ai;ai=ai+1;ai+1=t; /交換兩個(gè)數(shù)的位置，使小數(shù)上浮 cout<<"the sorted numbers ："<<endl; for(i=0;i<10;i+) /輸出10個(gè)數(shù) cout<<ai<<" " cout<<endl; return 0;例54將二維數(shù)組行列

22、元素互換，存到另一個(gè)數(shù)組中a=1 2 3 5 6 b=1 42 53 6#include <iostream>using namespace std;int main() int a23=1,2,3,4,5,6; int b32,i,j; cout<<"array a:"<<endl; for(i=0;i<=1;i+) for(j=0;j<=2;j+) cout<<aij<< " " bji=aij;cout<<endl; cout<<"array

23、b:“<<endl; for(i=0;i<=2;i+) for(j=0;j<=1;j+) cout<<bij<< " " cout<<endl; return 0;例55求 3×4 二維數(shù)組中最大元素值及其行列號(hào)#include <iostream>using namespace std;int main() int i,j,row=0,colum=0,max; int a34=5,12,23,56,19,28,37,46,-12,-34,6,8; max=a00; /使max開始時(shí)取a00的

24、值 for(i=0;i<=2;i+) /從第0行第2行for(j=0;j<=3;j+) /從第0列第3列if(aij>max) /如果某元素大于max max=aij; /max將取該元素的值row=i; /記下該元素的行號(hào)icolum=j; /記下該元素的列號(hào)jcout<<"max= "<<max<< ",row="<<row<< ",colum="<<colum<<endl;return 0;例5.7用選擇法對(duì)數(shù)組中10個(gè)整數(shù)按由

25、小到大排序#include <iostream>using namespace std;int main( )void select_sort(int array,int n); /函數(shù)聲明int a10,i; cout<<"enter the originl array："<<endl;for(i=0;i<10;i+) /輸入10個(gè)數(shù) cin>>ai; cout<<endl; select_sort( a ,10); /函數(shù)調(diào)用，數(shù)組名作實(shí)參 cout<<"the sorted arr

26、ay："<<endl; for(i=0;i<10;i+) /輸出10個(gè)已排好序的數(shù) cout<<ai<<" " cout<<endl; return 0;void select_sort(int array,int n) /形參array是數(shù)組名 int i,j,k,t; for(i=0;i<n-1;i+) k=i; for(j=i+1;j<n;j+) if(arrayj<arrayk) k=j; t=arrayk;arrayk=arrayi;arrayi=t; 習(xí)題：6、打印出楊輝三角（要求

27、打印出10行）#include <iostream>#include <iomanip>using namespace std;int main() const int n=11; int i,j,ann; for (i=1;i<n;i+) aii=1; ai1=1; for (i=3;i<n;i+) for (j=2;j<=i-1;j+) aij=ai-1j-1+ai-1j; for (i=1;i<n;i+) for (j=1;j<=i;j+) cout<<setw(5)<<aij<<" &q

28、uot; cout<<endl; cout<<endl; return 0;C語言部分1. 編程計(jì)算1, 1+2, 1+2+3, , 1+2+3+n各項(xiàng)值的積答案：#include <iostream>using namespace std;void main()int i,n,mul=1,sum=0;cout<<"Please input n:"cin>>n;for (i=1;i<=n;i+)sum+=i;mul*=sum;cout<<"The multiple is:"&l

29、t;<mul<<endl;2. 編寫重載函數(shù)Max可分別求取兩個(gè)整數(shù)，三個(gè)字符，兩個(gè)浮點(diǎn)型數(shù)，三個(gè)雙精度數(shù)的最大值。答案：int max(int i,int j)if (i>j)return i;else return j;char max(char c1,char c2,char c3)char ch;if (c1>c2)ch=c1;else ch=c2;if (c3>ch)ch=c3;return ch;float max(float x,float y)if (x>y)return x;else return y;double max(doubl

30、e x1,double x2,double x3)double y;if (x1>x2)y=x1;else y=x2;if (x3>y)y=x3;return y;3.請(qǐng)編寫一個(gè)函數(shù)int find(char s,char t)，該函數(shù)在字符串s中查找字符串t，如果找到，則返回字符串t在字符串s中的位置（整數(shù)值）；否則返回-1。注意：用數(shù)組方式及兩重循環(huán)來實(shí)現(xiàn)該函數(shù)。答案：int find(char s,char t)int i,j,slen,tlen;slen=strlen(s);tlen=strlen(t); i=0;j=0;while (i<slen &&

31、; j<tlen)if (si=tj)i+;j+;else i=i-j+1;j=0;if (j>=tlen)return i-tlen+1;else return -1;4.寫一程序，將字符串a(chǎn)的所有字符傳送到字符串b中，要求每傳送3個(gè)字符后放一個(gè)空格，例如，字符串a(chǎn)為”abcdefg”，則b為”abc deg g”。答案：#include <stdio.h>#include <string.h>#define N 40void fun(char s,char t) int i=0,j=0; while(si) tj=si; i+; j+; if (i)%3

32、=0) tj=' ' j+; tj='0'main() char aN,bN; printf("請(qǐng)輸入原始字符串a(chǎn): "); gets(a); fun(a,b); printf("加入空格后的字符串b: "); puts(b);5. 編程判斷一個(gè)數(shù)是否為素?cái)?shù)。答案：#include <iostream>#include <math.h>using namespace std;int prime(int m)int i,flag=1;for (i=2;i<=int(sqrt(m);i+)if (

33、m%i=0)flag=0;break; return flag;int main(void)int n;cout<<"請(qǐng)輸入一個(gè)整數(shù):"cin>>n;if (prime(n)cout<<n<<"是素?cái)?shù)!"<<endl;elsecout<<n<<"不是素?cái)?shù)!"<<endl;return 0;6.編程打印九九乘法表： 1×1=1 1×2=2 2×2=4 1×3=3 2×3=6 3×3=

34、9 .答案：#include <iostream>using namespace std;void main()int i,j;for (i=1;i<=9;i+)for (j=1;j<=i;j+)printf("%1d*%1d=%2d ",j,i,i*j); printf("n");7.請(qǐng)編寫一個(gè)程序，程序的功能是打印以下圖案。 * * * * * *答案：#include <iostream>using namespace std;void main()int i,j;for (i=1;i<=4;i+)for

35、(j=1;j<=4-i;j+)cout<<' 'for (j=1;j<=2*i-1;j+)cout<<'*'cout<<endl;for (i=3;i>=1;i-)for (j=1;j<=4-i;j+)cout<<' 'for (j=1;j<=2*i-1;j+)cout<<'*'cout<<endl;8. 寫出下面程序的運(yùn)行結(jié)果：#include <iostream.h>void average(float*p, in

36、t n)int i;float sum=0;for(i=0;i<n;i+,p+)sum+=(*p);cout<<sum/n;void main()float score22= 1,2,3,4 ;average(*score, 4);答案：2.5C+部分1定義一個(gè)滿足如下要求的Date類。（1）用下面的格式輸出日期：日/月/年（2）可運(yùn)行在日期上加一天操作；（3）設(shè)置日期。答案：#include <iostream> using namespace std; extern int i = 0; class Date public: set(int y,int m,

37、int d); add(); print(); private: int year; int month; int day; ; Date:set(int y, int m, int d) year = y; month = m; day = d; Date:add() switch(month) case 1: case 3: case 5: case 7: case 8: case 10: if(day < 31) day += 1; else if(day = 31) day = 1; month += 1; break; case 12: if(day < 31) day

38、+= 1; else if(day = 31) day = 1; month = 1; year += 1; break; case 4: case 6: case 9: case 11: if(day < 30) day += 1; else if(day = 30) day = 1; month += 1; break; case 2: if(year % 4 = 0 && year % 100 != 0 | year % 400 = 0) if(day < 29) day += 1; else if(day = 29) day = 1; month+= 1;

39、else if(day < 28) day += 1; else if(day = 28) day = 1; month += 1; break; Date:print() cout << day << "/" << month << "/" << year << endl; void main() Date d1; int year; int month; int day; cout << "請(qǐng)輸入年月日：" cin >> year

40、 >> month >> day; d1.set(year, month, day); d1.print(); d1.add(); cout << "下一天是：" ; d1.print(); 2定義一個(gè)時(shí)間類Time，能提供和設(shè)置由時(shí)、分、秒組成的時(shí)間，并編出應(yīng)用程序，定義時(shí)間對(duì)象，設(shè)置時(shí)間，輸出該對(duì)象提供的時(shí)間。答案：/Time.h class Time public: Time();Time();bool SetTime(int h,int m,int s); void GetTime(int &h,int &m,in

41、t &s); void Print(); private: int iHour,iMinute,iSecond;/*/Time.cpp#include <iostream>#include "Time.h" using namespace std;Time:Time()iHour=1;iMinute=1;iSecond=1;Time:Time()bool Time:SetTime(int h,int m,int s)if(h<0)|(h>23) return false; if(m<0)|(m>59) return false;

42、if(s<0)|(s>59) return false; iHour =h;iMinute= m;iSecond =s;return true; void Time:GetTime(int &h,int &m,int &s)h=iHour;m=iMinute;s=iSecond;void Time:Print()cout<<iHour<<':'<<iMinute<<':'<<iSecond<<endl; /*/mainfunction.cpp#includ

43、e <iostream>#include "Time.h"using namespace std;void main() Time t; t.SetTime(23,44,59);t.Print();3. 編寫一個(gè)矩形的類，要求類中具有長(zhǎng)和寬2個(gè)數(shù)據(jù)成員，具有構(gòu)造函數(shù)、計(jì)算矩形的面積即及周長(zhǎng)這3個(gè)成員函數(shù)。答案：class rectanglefloat width;float heigth;public:rectangle(float w, float h)width=w;heigth=h;float area()return width*heigth;float

44、 circulm()return 2*(width+heigth);4. 設(shè)計(jì)一個(gè)人的類，要求包含姓名、身份證號(hào)、年齡、以及人的總數(shù)和所有人的平均年齡這5個(gè)數(shù)據(jù)成員，然后編寫一個(gè)測(cè)試程序，輸入n個(gè)人的信息（姓名、身份證號(hào)和年齡），要求輸出這些人的信息和平均年齡。 答案：#include<iostream>#include<cstring>using namespace std;const int n=3;class Person private: char name20; char id18; int age; public: static int count; /人數(shù)

45、 static float aveage; Person(void) Person(const char* names,char *ids,int ages) strcpy(name,names);/初始化姓名 strcpy(id,ids); age = ages;/初始化年齡 aveage+=age; count+; void display() cout<<"姓名："<<name<<endl; cout<<"身份證："<<id<<endl; cout<<"年

46、齡："<<age<<endl; friend void ModifyAveage();/修改平均年齡的友元函數(shù);int Person:count = 0;float Person:aveage=0;void ModifyAveage()Person:aveage=Person:aveage/Person:count;void main() int ages,i; char namestr20,idstr18; Person pern; for (i=0;i<n;i+) cout<<"請(qǐng)輸入姓名：" cin>>n

47、amestr; cout<<"請(qǐng)輸入身份證號(hào)碼：" cin>>idstr; cout<<"請(qǐng)輸入的年齡：" cin>>ages; Person temp(namestr,idstr,ages); peri=temp; ModifyAveage(); for (i=0;i<n;i+) peri.display(); cout<<"人的數(shù)量："<<Person:count<<endl; cout<<"平均年齡："<

48、;<Person:aveage<<endl;5. 編寫一個(gè)圓的類，再使用派生類的機(jī)制，編寫一個(gè)圓柱的類，求每個(gè)類都包含具有初始化功能的構(gòu)造函數(shù)和相關(guān)數(shù)據(jù)的輸出函數(shù)。答案：#include <iostream>#define pi 3.1416using namespace std; class circle protected: float r; float area; public: circle() r=0; circle(float radius)r=radius; void print() area=r*r*pi; cout<<"r=

49、"<<r<<" area="<<area<<endl; ; class column:circle protected: float h; float v; public : column(float radius,float height):circle(radius) h=height; void print() circle:print(); cout<<"h="<<h<<" volume="<<area*h<<

50、endl; ; void main() column col (3.0,2.0); col.print(); 6.重載一個(gè)函數(shù)模板，要求它可以返回兩個(gè)值中的較大者，也可以求三個(gè)數(shù)的最大值。答案：/ 求兩個(gè)任意類型值中的最大者template <typename T>inline T const& max (T const& a, T const& b) return a < b ? b : a;/ 求3個(gè)任意類型值中的最大者template <typename T>inline T const& max (T const&

51、a, T const& b, T const& c) return :max (:max(a,b), c);7.下面是整型安全數(shù)組類的一部分：試完善類的定義，使下段程序能夠正常運(yùn)行且不會(huì)出現(xiàn)內(nèi)存泄漏。#include <iostream.h>class Arraypublic:Array(int x)count=x;p=new intcount;int & operator (int x)return *(p+x);protected:int count;int *p;答案：#include <iostream.h>class Arraypubl

52、ic:Array(int x); Array(Array &); Array(); void SetAt(int pos, int value); int GetAt(int pos); protected:int count;int *p;Array:Array(int x)count=x;p=new intcount;Array:Array(Array &a)int i;count=a.count;p=new intcount;for (i=0;i<count;i+) *(p+i)=*(a.p+i);Array:Array()delete p;void Array:SetAt(i