课后作业，帮忙编写完整的C++程序

答案：水仙花束的：
#include <iostream>
using namespace std;
int main()
{
 int n=0;
 int f,s,t;
 cout<<"水仙花数为:\n";
 for(n=100;n<1000;n++)
 {
  f=n/100;
  s=n/10-f*10;
  t=n-f*100-s*10;
  if(f*f*f+s*s*s+t*t*t==n)
   cout<<n<<endl;
 }
 return 0;
}
 
定义一个数组，数据为：6，3，7，1，4，8，2，9，11，5，创建一个向量，把数组赋给该向量，然后求该向量的标准差。
#include<iostream>
#include<vector>
#include<cmath>
using namespace std;
int main()
{
 int a[10]={6,3,7,1,4,8,2,9,11,5};
 int i;;
 double s,r=0,average=0;
 vector<int> x(a,a+10);
 for (i=0;i<10;i++){
  average+=x[i]/10;
     r+=x[i]-pow(average,2);
 }
 s=sqrt(r/10);
 cout<<s<<endl;
}
 
打印99乘法表
#include "stdio.h"
#include "math.h"
int main(void)
{
    int flag, i, j, n;
    int a[6][6];
    int repeat, ri;
    scanf("%d", &repeat);
    for(ri = 1; ri <= repeat; ri++){
        scanf("%d", &n);
        for(i = 0; i < n; i++)
            for(j = 0; j < n; j++)
                scanf("%d", &a[i][j]);
         for(i=0;i<n;i++)
    for(j=0;j<i;j++){
     if(a[i][j]!=0)
      flag=0;
    }
        if(flag != 0)  printf("YES\n");
        else  printf("NO\n");
    }
}
编程实现输入一个整数，输出相应的五分制成绩，90分及以上为’A’，80~89为’B’……59分及以下为’E‘
#include <iostream>
#include <ostream>
using namespace std;
int main()
{
 int grade;
 cin>>grade;
 if (grade>=90)
  grade=1;
 else if (grade>=80&&grade<90)
  grade=2;
 else if (grade>=70&&grade<80)
  grade=3;
 else if (grade>=60&&grade<70)
  grade=4;
 else 
  grade=5;
 switch(grade)
 {
 case 1:cout<<"A"<<endl;break;
 case 2:cout<<"B"<<endl;break;
 case 3:cout<<"C"<<endl;break;
 case 4:cout<<"D"<<endl;break;
 case 5:cout<<"E"<<endl;break;
 }
 return 0;
}
 
打印整数-1234567的二进制位码
#include<iostream>

using namespace std;

int main()
{
int a=-123456,i;
for(i=31;i>=0;i--)
cout<<(a>>i&1);
cout<<endl; 
return 0;
} 
 
编程求值，注意不要数据溢出
#include<iostream>

using namespace std;

int main()
{
double factorial(int n);
double result=0;
result=factorial(18)/(factorial(13)*factorial(5));
cout<<result<<endl;
} 

double factorial(int n)
{
int i;
double sum=1;

if(n==0) 
sum=0;

for(i=1;i<=n;i++)
sum*=i;
return sum;
}
用递归算法实现如下问题：球从100米高空坠落，落地后反弹原高度的一半，求第10次落地后反弹的高度和已经过的距离
#include<iostream>
using namespace std;
int main()
{
 void freedown(int &n,double &rhight,double &distence);
    int n=10;
 double rhight=100;
 double distence=0;
 freedown(n,rhight,distence);
 cout<<"第十次落地后反弹的高度："<<rhight<<endl<<"经过的距离："<<distence;
} 
void freedown(int &n,double &rhight,double &distence)
{
 
 while(n){
 rhight/=2;
 distence+=3*rhight;
 freedown(--n,rhight,distence);
 }
}
 
实现堆栈和回文判断 
#include<iostream> #include<cstdlib>using namespace std;#define MaxSize 100 #define ElemType char typedef struct { ElemType data[MaxSize]; int top; }SeqStack; void StackInitial(SeqStack *pS) { pS->top=-1; } int IsEmpty(SeqStack *pS) { return pS->top==-1; } int IsFull(SeqStack *pS) { return pS->top>=MaxSize-1; } void Push(SeqStack *pS,ElemType e) { if(IsFull(pS)) { cout<<"Full"; exit(1); } pS->data[++pS->top]=e; } ElemType Pop(SeqStack *pS) { if(IsEmpty(pS)) { cout<<"Empty"; exit(1); } return pS->data[pS->top--]; } ElemType GetTop(SeqStack *pS) { if(IsEmpty(pS)) { cout<<"Empty"; exit(1); } return pS->data[pS->top]; } void MakeEmpty(SeqStack *pS) { pS->top=-1; } void main() {ElemType d1[100],ch; int i=0,j=0;//,a; /////////////////SeqStack stack; StackInitial(&stack); cout<<"Please input a character string\n"; cin>>d1; while((ch=d1[i++])!='\0') { Push(&stack,ch); } while(!IsEmpty(&stack)) { if(Pop(&stack)!=d1[j++])/////////////;cout<<"no"; } cout<<"yes"; }  
编程实现复数类和实数类，实数类以复数类的子类来实现。复数类要求实现操作符重载，
#include<iostream>#include<cstdlib>#include<cmath>using namespace std;class complex{private:    double re;    double im;public:complex(){}        complex(double a,double b){re=a;im=b;}    void set(double a,double b){re=a;im=b;} double geti() {return im; } double getr() {return re; }    complex operator +=(complex &a){    this->re+=a.re;        this->im+=a.im;        return *this;    }    complex operator -=(complex &a){    this->re-=a.re;        this->im-=a.im;        return *this;    }    complex operator *=(complex &a){    re =re*a.re-im*a.im;        im =re*a.im+a.re*im;        }    complex  operator /=(complex &a)    {    if(a.re||a.im) {    re =(re*a.re+im*a.im)/(a.re*a.re+a.im*a.im);            im =(re*a.im-a.re*im)/(a.re*a.re+a.im*a.im); }    }    friend ostream& operator << (ostream& os, complex& c);    friend istream& operator >> (istream& is, complex& c);friend complex operator +(complex &a, complex &b);    friend complex operator -(complex &a);     friend complex operator -(complex &a, complex &b);    friend complex operator *(complex &a, complex &b);    friend complex operator /(complex &a, complex &b);    friend int operator ==(complex &a, complex &b);     friend int operator !=(complex &a, complex &b);};complex operator +(complex &a, complex &b){     complex  temp;     temp.re = a.re + b.re;     temp.im = a.im + b.im;     return temp;}complex operator -(complex &a){a.re=-a.re;a.im=-a.im;return a;}complex operator -( complex &a, complex &b ){     complex temp;     temp.re = a.re - b.re;     temp.im = a.im - b.im;     return temp;}complex operator *(complex &a, complex &b){complex temp;temp.re = a.re * b.re - a.im * b.im;temp.im = a.re * b.im + a.im * b.re;return temp;}complex operator /(complex &a, complex &b){complex temp;temp.re = (a.re * b.re + a.im * b.im) / (b.re * b.re + b.im * b.im);temp.im = (a.im * b.re - a.re * b.im) / (b.re * b.re + b.im * b.im);return temp;} int operator ==(complex &a, complex &b){return (a.re == b.re && a.im == b.im);}int operator !=(complex &a, complex &b){return !(a==b);} ostream& operator << (ostream& os, complex& c) { os << c.re;     if(c.im > 0)        os << "+" << c.im &
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