欧美日韩视频不卡,国产午夜精品久久久久久久,欧美日韩成人精品

mahudu@cppblog — Wed, 28 Jun 2006 03:42:00 GMT

在水木上见到一个脓子，关于虚函数的讉K��权限问题�Q?br />

#include<cstdlib>

#include<iostream>

using namespace std;

class B {

public:

   virtual int f1(){cout<<"B::f1()"<<endl;return 0;}

   virtual void f2( int val){cout<<"B::f2(int)"<<endl;}

   virtual int f3( int val ){cout<<"B::f3(int)"<<endl;return 0;}

};

class D : public B {

   int f1(){cout<<"D::f1()"<<endl;return 0;}

   virtual void f4(){cout<<"D::f4()"<<endl;}

   int f3(int val){cout<<"D::f3(int)"<<endl;return 0;}

};

int main(int argc, char *argv[])

{

   B *bp = new D;

   bp->f3(12);//D中的f3是private的，可以讉K��#1

   D *dp=new D;

   dp->f3(12);//f3是private,讉K��不了�Q�编译通不�q?/span>

   system("PAUSE");

   return EXIT_SUCCESS;

}

其实�q�是一个关于访问权限决定时间的问题�Q�由�?u>讉K��权限是编译时间决定的�Q�而不是运行时军_��的�?br />B *bp = new D; // 此时bp所指向的类型是B而不是D�Q�而B的f3()是公有的�Q�所以可以访问�?br />D *dp = new D; // 此时dp所指向的类型是D�Q�而D的f3()是私有的�Q�所以不能访问�?br />

mahudu@cppblog 2006-06-28 11:42 发表评论

Self Numbers

mahudu@cppblog — Fri, 16 Jun 2006 15:32:00 GMT

In 1949 the Indian mathematician D.R. Kaprekar discovered a class

of numbers called self-numbers. For any positive integer n, define

d(n) to be n plus the sum of the digits of n. (The d stands

for digitadition, a term coined by Kaprekar.) For example, d(75) = 75 + 7 + 5 = 87. Given any positive integer n as a starting

point, you can construct the infinite increasing sequence of integers

n, d(n), d(d(n)), d(d(d(n))), .... For example, if you start with

33, the next number is 33 + 3 + 3 = 39, the next is 39 + 3 + 9 = 51, the next

is 51 + 5 + 1 = 57, and so you generate the sequence

33, 39, 51, 57, 69, 84, 96, 111, 114, 120, 123, 129, 141, ...

The number n is called a generator of d(n). In the

sequence above, 33 is a generator of 39, 39 is a generator of 51, 51

is a generator of 57, and so on. Some numbers have more than one

generator: for example, 101 has two generators, 91 and 100. A number

with no generators is a self-number. There are thirteen

self-numbers less than 100: 1, 3, 5, 7, 9, 20, 31, 42, 53, 64, 75, 86,

and 97.

Write a program to output all positive self-numbers less than 10000

in increasing order, one per line.

Output
1
3
5
7
9
20
31
42
53
64
 |
 |       <-- a lot more numbers
 |
9903
9914
9925
9927
9938
9949
9960
9971
9982
9993

Solution

#include <iostream>
using namespace std;

const long N = 10000;     //最大自然数
char Arr[N + 9*4]={0};   //是否是被排除的数字？ +9*4是�ؓ了要再多�?位数

long DealNum(long n)
{
  long sum = n;
  while (n != 0)
  {
    sum += n%10;
    n /= 10;
  }
  return sum;
}

int main()
{
  int i;
  for(i = 1; i < N; i++)
  {
    Arr[DealNum(i)] = 1;
  }
  for(i = 1; i < N; i++)
  {
    if (!Arr[i])
        cout<<i<<endl;
  }
  return 0;
}

mahudu@cppblog 2006-06-16 23:32 发表评论

Booklet Printing

mahudu@cppblog — Fri, 16 Jun 2006 15:26:00 GMT

When printing out a document, normally the first page is printed first, then the second, then the third, and so on until the end. However, when creating a fold-over booklet, the order of printing must be altered. A fold-over booklet has four pages per sheet, with two on the front and two on the back. When you stack all the sheets in order, then fold the booklet in half, the pages appear in the correct order as in a regular book.

For example, a 4-page booklet would print on 1 sheet of paper: the front will contain page 4 then page 1, and the back will contain page 2 then page 3.

                       Front              Back
                       -------------      -------------
                       |     |     |      |     |     |
                       |  4  |  1  |      |  2  |  3  |
                       |     |     |      |     |     |
                       -------------      -------------

Your task is to write a program that takes as input the number of pages to be printed, then generates the printing order.

Input

The input file contains one or more test cases, followed by a line containing the number 0 that indicates the end of the file.

Each test case consists of a positive integer n on a line by itself, where n is the number of pages to be printed; n will not exceed 100.

Output

For each test case, output a report indicating which pages should be printed on each sheet, exactly as shown in the example. If the desired number of pages does not completely fill up a sheet, then print the word Blank in place of a number. If the front or back of a sheet is entirely blank, do not generate output for that side of the sheet.

Output must be in ascending order by sheet, front first, then back.

Sample Input

Sample Output

Printing order for 1 pages:
Sheet 1, front: Blank, 1
Printing order for 14 pages:
Sheet 1, front: Blank, 1
Sheet 1, back : 2, Blank
Sheet 2, front: 14, 3
Sheet 2, back : 4, 13
Sheet 3, front: 12, 5
Sheet 3, back : 6, 11
Sheet 4, front: 10, 7
Sheet 4, back : 8, 9
Printing order for 4 pages:
Sheet 1, front: 4, 1
Sheet 1, back : 2, 3

Solution

#include <iostream>
using namespace std;
#define PAGES 100

typedef struct side{    
    int left,right;
}side;

typedef struct sheet{
    side front;
    side back;    
}sheet;

int numSides;
sheet sheets[PAGES];

void PrintPages(int numSides){
    int numSidesNew;    
    int add,pages;
    add = numSides%4;
    if(add != 0){
        numSidesNew = numSides + 4 - add;    // 增加后的总面�?numSides为实际的总面�?/span>
    }
    else
        numSidesNew = numSides;
    pages = numSidesNew / 4;    // �ȝ��张数
    for(int i = 0; i < pages; i++){
        sheets[i].front.left = numSidesNew - 2*i;
        if(sheets[i].front.left > numSides){
            sheets[i].front.left = 0;    // 表明应�ؓblank
        }
        sheets[i].front.right = 2*i+1;
        if(sheets[i].front.right > numSides){
            sheets[i].front.right = 0;    // 表明应�ؓblank
        }
        sheets[i].back.left = 2*(i+1);
        if(sheets[i].back.left > numSides){
            sheets[i].back.left = 0;    // 表明应�ؓblank
        }
        sheets[i].back.right = numSidesNew - 2*i - 1;
        if(sheets[i].back.right > numSides){
            sheets[i].back.right = 0;
        }
    }

    cout << "Printing order for " << numSides << " pages:" << endl;
    for(int j = 0; j < pages; j++){
        if(sheets[j].front.left || sheets[j].front.right){
            cout << "Sheet " << j+1 <<", front: ";
            if(sheets[j].front.left)
                cout << sheets[j].front.left << ",";
            else
                cout << "Blank,";
            cout << " ";
            if(sheets[j].front.right)
                cout << sheets[j].front.right;
            else
                cout << "Blank,";
            cout << endl;
        }
        if(sheets[j].back.left || sheets[j].back.right){
            cout << "Sheet " << j+1 <<", back : ";
            if(sheets[j].back.left)
                cout << sheets[j].back.left << ",";
            else
                cout << "Blank,";
            cout << " ";
            if(sheets[j].back.right)
                cout << sheets[j].back.right;
            else
                cout << "Blank";
            cout << endl;
        }

    }
}


int main()
{
    int numSides;
    while(cin >> numSides){
        if(numSides == 0){
            break;
        }
        PrintPages(numSides);
    }
    return 0;
}

mahudu@cppblog 2006-06-16 23:26 发表评论

The Blocks Problem

mahudu@cppblog — Fri, 09 Jun 2006 17:16:00 GMT

摘要: Background Many areas of Computer Science use simple, abstract domains for both analytical and empirical studies. For example, an early AI study of planning and robotics (STRIPS) used a block worl... 阅读全文

mahudu@cppblog 2006-06-10 01:16 发表评论

Fibonacci Freeze

mahudu@cppblog — Fri, 09 Jun 2006 17:10:00 GMT

The Fibonacci numbers (0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, ...) are defined by the recurrence:

Write a program to calculate the Fibonacci Numbers.

Input and Output

The input to your program would be a sequence of numbers smaller or equal than 5000, each on a separate line, specifying which Fibonacci number to calculate.

Your program should output the Fibonacci number for each input value, one per line.

Sample Input

5
7
11

Sample Output

The Fibonacci number for 5 is 5
The Fibonacci number for 7 is 13
The Fibonacci number for 11 is 89

Solution

#include

using namespace std;

int main()

{

int first,next,temp,n;

while(cin >> n) {

first = 0;

next = 1;

temp = 0;

if(n == 0 || n == 1) {

cout << "The Fibonacci number for" << " " << n << " " << "is" << " " << n << endl;

}

else {

for(inti = 2; i <= n; i++) {

temp = first + next;

first = next;

next = temp;

}

cout << "The Fibonacci number for" << " " << n << " " << "is" << " " << temp << endl;

}

return 0;

}

mahudu@cppblog 2006-06-10 01:10 发表评论

The 3n + 1 problem

mahudu@cppblog — Fri, 09 Jun 2006 16:41:00 GMT

Background

Problems in Computer Science are often classified as belonging to a certain class of problems (e.g., NP, Unsolvable, Recursive). In this problem you will be analyzing a property of an algorithm whose classification is not known for all possible inputs.

The Problem

Consider the following algorithm:

1.	input n

2. 	print n

3. 	if n = 1 then STOP

4. 	if n is odd then  tex2html_wrap_inline44

5. 	else  tex2html_wrap_inline46

6. 	GOTO 2

Given the input 22, the following sequence of numbers will be printed 22 11 34 17 52 26 13 40 20 10 5 16 8 4 2 1

It is conjectured that the algorithm above will terminate (when a 1 is printed) for any integral input value. Despite the simplicity of the algorithm, it is unknown whether this conjecture is true. It has been verified, however, for all integers n such that 0 < n < 1,000,000 (and, in fact, for many more numbers than this.)

Given an input n, it is possible to determine the number of numbers printed (including the 1). For a given n this is called the cycle-length of n. In the example above, the cycle length of 22 is 16.

For any two numbers i and j you are to determine the maximum cycle length over all numbers between i and j.

The Input

The input will consist of a series of pairs of integers i and j, one pair of integers per line. All integers will be less than 1,000,000 and greater than 0.

You should process all pairs of integers and for each pair determine the maximum cycle length over all integers between and including i and j.

You can assume that no opperation overflows a 32-bit integer.

The Output

For each pair of input integers i and j you should output i, j, and the maximum cycle length for integers between and including i and j. These three numbers should be separated by at least one space with all three numbers on one line and with one line of output for each line of input. The integers i and j must appear in the output in the same order in which they appeared in the input and should be followed by the maximum cycle length (on the same line).

Sample Input

Sample Output

1 10 20
100 200 125
201 210 89
900 1000 174

Solution

#include

using namespace std;

int cycle(intm)

{

int i = 1;

while (m != 1){

if(m%2)

m = m*3 + 1;

else

m /= 2;

i++;

}

return i;

}

int main()

{

int m,n,max,temp;

int mOriginal,nOriginal;

int i;

while (cin >> m >> n){

mOriginal = m;

nOriginal = n;

if (m > n){

temp = m;

m = n;

n = temp;

}

max = cycle(m);

for (i = m+1; i <= n; i++){

temp = cycle(i);

if (temp > max){

max = temp;

}

cout << mOriginal << " " << nOriginal << " " << max << endl;

}

return 0;

}

mahudu@cppblog 2006-06-10 00:41 发表评论

A + B Problem

mahudu@cppblog — Fri, 09 Jun 2006 16:32:00 GMT

Calculate a + b

Input

a and b

Output

a+b

Sample Input

1 5

Sample Output

Solution

#include
using namespace std;

int main()
{
     int a,b;
    while(cin >> a >> b);
         cout << a+b << endl;
     return 0;
}

mahudu@cppblog 2006-06-10 00:32 发表评论