求2的n次方～

wanglei — Fri, 08 Jun 2007 19:01:00 GMT

   /*----------------------------
    ×××求2的n次方的程序×××
   ----------------------------*/
#include
#include
#include
using namespace std;

void f(int n)
{
if(n)
{
  const int number = 2;
  vector vec(1,number);
  int i = 1;
  while(i != n)
  {
   vector::iterator iter = vec.begin();
   vector::iterator itd = vec.begin();
   for(;itd != vec.end(); ++ itd)
    (*itd) *= 2;
   for(;iter != vec.end(); ++ iter)
   {
    vector::iterator iter_ = iter + 1;
    if((iter_) != vec.end())
    {
     if((*iter) > 10)
     {
      (*iter) = (*iter) % 10;
      (*(iter + 1)) ++;
     }
    }
    else
     if((*iter) >= 10)
     {
      (*iter) %= 10;
      vec.push_back(1);
      break;
     }
   }
   ++i;
  }
  vector::iterator j = vec.end() - 1;
  while(j != vec.begin())
  {
   cout << (*j);
   j--;
  }
  j = vec.begin();
  cout << (*j);
  return;
}
cout << "1\n";
return;
}

int main()
{
int n;
cout << "Input the number:";
cin >> n;
cout << "The rusult is:";
f(n);
system("PAUSE");
return 0;
}

wanglei 2007-06-09 03:01 发表评论

输出如下的一个矩阵～

wanglei — Fri, 08 Jun 2007 18:57:00 GMT

/*
   --------------------------------------------
   输入一个矩阵的宽度N,（如N=3）输出如下格式：
          1 2 3
          8 9 4
          7 6 5
   --------------------------------------------
*/
#include
#include
using namespace std;
namespace wl{
int i,j;
static int iter = 1;
int* q;
int n;
}
int* putarr(int **p,int lhs,int rhs)
{ //lhs为行数，rhs为列数
using namespace wl;
if(i == 0)
{
    if(j != (rhs-1))
    {
      p[i][j+1] = iter;
      iter ++;
      j++;
      return q = &p[i][j+1];
    }
    p[i+1][j] = iter;
    iter ++;
    i++;
    return q = &p[i+1][j];
}
else if(j ==(rhs-1))
{
    if(i != (lhs-1))
    {
     p[i+1][j] = iter;
     iter ++;
     i ++;
     return q = &p[i+1][j];
    }
    p[i][j-1] = iter;
    iter ++;
    j--;
    return q = &p[i][j-1];
}
else if(i ==(lhs-1))
{
    if(j != 0)
    {
      p[i][j-1] = iter;
      iter ++;
      j--;
      return q = &p[i][j-1];
    }
    p[i-1][j] = iter;
    iter ++;
    i--;
    return q = &p[i-1][j];
}
else if(j == 0)
{
    if(i != 0)
    {
      if(p[i-1][j] == 0)
      {
     p[i-1][j] = iter;
     iter ++;
     i--;
     return q = &p[i-1][j];
      }
      p[i][j+1] = iter;
      iter ++;
      j++;
      return q = &p[i][j+1];
    }
    p[i][j+1] = iter;
    iter ++;
    j++;
    return q = &p[i][j+1];
}
else
{
    if((p[i-1][j] == 0)&&(p[i][j-1] != 0)&&(p[i+1][j] != 0)&&(p[i][j+1] != 0))
    {
      p[i-1][j] = iter;
      iter ++;
      i--;
      return q = &p[i-1][j];
    }
    else if((p[i-1][j] != 0)&&(p[i][j-1] == 0)&&(p[i+1][j] != 0)&&(p[i][j+1] != 0))
    {
      p[i][j-1] = iter;
      iter ++;
      j--;
      return q = &p[i][j-1];
    }
    else if((p[i-1][j] != 0)&&(p[i][j-1] != 0)&&(p[i+1][j] == 0)&&(p[i][j+1] != 0))
    {
      p[i+1][j] = iter;
      iter ++;
      i++;
      return q = &p[i+1][j];
    }
    else if((p[i-1][j] != 0)&&(p[i][j-1] != 0)&&(p[i+1][j] != 0)&&(p[i][j+1] == 0))
    {
      p[i][j+1] = iter;
      iter ++;
      j++;
      return q = &p[i][j+1];
    }
    else if((p[i-1][j] == 0)&&(p[i][j+1] == 0))
    {
      p[i-1][j] = iter;
      iter ++;
      i--;
      return q = &p[i-1][j];
    }
    else if((p[i][j+1] == 0)&&(p[i+1][j] == 0))
    {
      p[i][j+1] = iter;
      iter ++;
      j++;
      return q = &p[i][j+1];
    }
    else if((p[i+1][j] == 0)&&(p[i][j-1] == 0))
    {
      p[i+1][j] = iter;
      iter ++;
      i++;
      return q = &p[i+1][j];
    }
    else if((p[i][j-1] ==0)&&(p[i-1][j] == 0))
    {
      p[i][j-1] = iter;
      iter ++;
      j--;
      return q = &p[i][j-1];
    }
    else return q = 0;
     }
}
int main()
{
using namespace wl;
i = 0;
j = 0;
cout << "输入数组的行数和列数n:" ;
cin >> n;
int** a = new int*[n];
for(int x = 0; x != n; ++x)
     a[x] = new int[n];
for(int x = 0; x != n; ++x)
     for(int y = 0; y != n; ++y)
        a[x][y] = 0;
a[0][0] = iter;
iter ++;
q = &a[0][0];
while(q&&(iter <= n*n))
    putarr(a,n,n);
for(int x = 0; x != n; ++x)
{
    for(int y = 0; y != n; ++y)
     cout << setw(3) << a[x][y] << " ";
    cout << '\n';
}
for(int k = n; k != 0; --k)
     delete []a[k-1];
delete []a;
system("PAUSE");
return 0;
}

wanglei 2007-06-09 02:57 发表评论

C++博客- WangLei's Blog-文章分类-C++代码

求2的n次方～

输出如下的一个矩阵～