我出一个，很简单的，Selection Algorithm,

该算法的功能是：1。Sorting the list in decreasing order, and then the Kth largest value will be in position K.

2。A related technique to this would be to find the largest value and then move it to the last location

in the list. If we again look for the largest value in the list, ignoring the value we already found,

we get the second largest value, which can be moved to the second last location in the list.

If we con tinue this process, we will find the Kth largest value on the Kth pass.

This gives the algorithm:

#include"iostream"

using namespace std;

void Find_Kth_Largest(int list[],int n,int k){

//list:the value to look through

//n:the size of the list

//k:the element to select

int i,largest,largest_location,j;

for(i=1;i<=k;i++){//for_1

largest=list[1];

largest_location=1;

for(j=2;j<=n-(i-1);j++){//for_2

if(list[j]>largest){

largest=list[j];

largest_location=j;

}//if

}//for_2

swap(list[n-(i-1)],list[largest_location]);//exchanging, making the largest elements

//in the last elements of the list

}//for_1

cout<<"The number is:"<<largest<<"\n";

}//Find_Kth_Largest

void swap(int &a,int &b){//exchange

int temp;

temp=a;

a=b;

b=temp;

}//swap

void main(){

int List[16],K,i;

cout<<"Please input the K:";

cin>>K;

cout<<"\n";

List[0]=0;

for(i=1;i<=15;i++){

List[i]=2*i-1;

List[0]++;//List[0] is stored the length of the List

}//for

cout<<"\n"<<List[0]<<"\n";//output the length

Find_Kth_Largest(List,List[0],K);

}

[此贴子已经被zinking于2005-10-21 13:44:26编辑过]

偶来个，，看上去挺吓人的，其实更简单～

/*【Background】

Personally it's considered as meaningful in exercise

【Problem Description】

求n之内的所有偶数表示为两个素数之和。 注：歌德巴赫猜想是要证明对任何大于6的自然数n命题成立。

【Input Example】

10

【Output】

6=3+3

8=3+5

10=3+7 */

file://IDE:VC++6.0 file://long型最大值范围内求解素数(之所以只是用long，因为范围再大也是一样永远是有限个数无法完全 file://证明此命题，只是一个示例验证罢了) file://素数就用的那经典判定法：2到此数square root不能被此数整除 #include <iostream.h> #include <math.h> #include <conio.h>

bool isPrime(long n) { for(int i=2;i<=sqrt(n);i++) if(n%i==0) return false; return true; }

int main() { long n; cin>>n; for(long i=6;i<=n;i+=2) for(long j=3;j<=i;j+=2) if(isPrime(j)&&isPrime(i-j)) { cout<<i<<"="<<j<<"+"<<i-j<<endl; break; } getch(); return 0; }

我水，就写个简单点的好了（复数——学习重载最好了）

#include<iostream.h> class Fushu { public: Fushu(){}; Fushu(float r,float i); friend Fushu operator +(Fushu &c1,Fushu &c2); friend Fushu operator -(Fushu &c1,Fushu &c2); friend Fushu operator *(Fushu &c1,Fushu &c2); friend Fushu operator /(Fushu &c1,Fushu &c2); void display(); private: float real; float imag; }; Fushu::Fushu(float r,float i) { real=r; imag=i; } void Fushu::display() { if(real!=0) { if(imag>0) { if(imag=1) cout<<real<<"+i"<<endl; else cout<<real<<"+"<<imag<<"i"<<endl; } if(imag<0) cout<<real<<imag<<"i"<<endl; else cout<<real<<endl; } else { if(imag>0) { if(imag=1) cout<<"i"<<endl; else cout<<"+"<<imag<<"i"<<endl; } if(imag<0) cout<<imag<<"i"<<endl; if(imag=0) cout<<"0"<<endl; } } Fushu operator +(Fushu &c1,Fushu &c2) { Fushu temp; temp.real=c1.real+c2.real; temp.imag=c1.imag+c2.imag; return temp; } Fushu operator -(Fushu &c1,Fushu &c2) { Fushu temp; temp.real=c1.real-c2.real; temp.imag=c1.imag-c2.imag; return temp; } Fushu operator *(Fushu &c1,Fushu &c2) { Fushu temp; temp.real=c1.real*c2.real-c1.imag*c2.imag; temp.imag=c1.real*c2.imag+c2.real*c1.imag; return temp; } Fushu operator /(Fushu &c1,Fushu &c2) { Fushu temp; float tt; tt=c2.real*c2.real+c2.imag*c2.imag; temp.real=(c1.real*c2.real+c1.imag*c2.imag)/tt; temp.imag=(-c1.real*c2.imag+c2.real*c1.imag)/tt; return temp; } int main() { Fushu x(1,1),y(1,-1),z; z=x+y; z.display(); z=x-y; z.display(); z=x*y; z.display(); z=x/y; z.display(); return 0; }

我把书上的Selection Algorithm编程实现了：

#include"iostream"

using namespace std;

void Find_Kth_Largest(int list[],int n,int k){

//list:the value to look through

//n:the size of the list

//k:the element to select

int i,largest,largest_location,j;

for(i=1;i<=k;i++){//for_1

largest=list[1];

largest_location=1;

for(j=2;j<=n-(i-1);j++){//for_2

if(list[j]>largest){

largest=list[j];

largest_location=j;

}//if

}//for_2

swap(list[n-(i-1)],list[largest_location]);//exchanging, making the largest elements

//in the last elements of the list

}//for_1

cout<<"The number is:"<<largest<<"\n";

}//Find_Kth_Largest

void swap(int &a,int &b){//exchange

int temp;

temp=a;

a=b;

b=temp;

}//swap

void main(){

int List[16],K,i;

cout<<"Please input the K:";

cin>>K;

cout<<"\n";

List[0]=0;

for(i=1;i<=15;i++){

List[i]=2*i-1;

List[0]++;//List[0] is stored the length of the List

}//for

cout<<"\n"<<List[0]<<"\n";//output the length

Find_Kth_Largest(List,List[0],K);

}

//N个小孩围成一圈报数，凡是报到指定数字的小孩离开圈子 //打印最后剩下的小孩的号码

#include<iostream.h>

void main() { const int N=10; //假定有10个小孩 int kid[N],*p; int interval; //报到此数的小孩离开 int count=0,leave=0; //报数计数器和离开的小孩数

for(int i=0;i<N;i++) kid[i]=i+1; //给每个小孩一个号码

cout<<"please input the Count off number: "; cin>>interval;

while(leave!=N-1) //当离开人数不等于总人数 { for(p=kid;p<kid+N;p++) //从号码是1的小孩开始数

if(*p!=0) //已离开的小孩不用报数 { count++;

if(count==interval) //报到此数时 { count=0; //由1开始重新报数 cout<<*p<<" "; *p=0; //离开的小孩号码置零 leave++; } } } ” 其中： “while(leave!=N-1) //当离开人数不等于总人数” 中注释是不是应该“//当离开人数不等于总人数‘减1啊？’”

[此贴子已经被作者于2005-10-18 16:20:44编辑过]