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列出所�?位数�Q�它的前n位能被n整除

Dain — Mon, 16 Apr 2007 09:29:00 GMT

最��单的是穷举，不过那可要O(9*10⁹)�Q�不可取

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

vector<int> fun(int n)
{
    vector<int> last,all;
    int i,j,k;
    for(i = 1;i < 10;++i)
        all.push_back(i);

    if(n == 1)
        return all;

    int size;
    int num;
    for(i = 2;i <= n;++i)
    {
        last = all;
        all.clear();
        size = (int)last.size();
        for(j = 0;j < size;++j)
        {
            for(k = 0;k < 10;++k)
            {
                num = last[j] * 10 + k;
                if(num % i == 0)
                    all.push_back(num);
            }
        }
        last.clear();
    }

    return all;
}

Dain 2007-04-16 17:29 发表评论

最大的子序列和问题

Dain — Wed, 07 Feb 2007 02:52:00 GMT

求解该问题的四种��法�Q?br /> 旉��O(N³)�Q?/em>��法一

int MaxSubsequenceSum( const int A[], int N)
{
     int ThisSum,MaxSum,i,j,k;

    MaxSum = 0 ;
     for (i = 0 ;i < N;i ++ )
         for (j = i;j < N;j ++ )
         {
            ThisSum = 0 ;
             for (k = i;k <= j;k ++ )    ThisSum += A[k];
             if (ThisSum > MaxSum)    MaxSum = ThisSum;
        }

     return MaxSum;
}
旉��O(N²)�Q?/em>��法�?/strong>

int MaxSubsequenceSum( const int A[], int N)
{
     int ThisSum,MaxSum,i,j;

    MaxSum = 0 ;
     for (i = 0 ;i < N;i ++ )
     {
        ThisSum = 0 ;
         for (j = i;j < N;j ++ )
         {
            ThisSum += A[k];
             if (ThisSum > MaxSum)    MaxSum = ThisSum;
        }
    }

     return MaxSum;
}
旉��O(NlogN)�Q�算法三

static int MaxSubSum( const int A[], int Left, int Right)
{
     int MaxLeftSum,MaxRightSum;
     int MaxLeftBorderSum,MaxRightBorderSum;
     int LeftBorderSum,RightBorderSum;
     int Center,i;

     if (Left == Right)
         if (A[left] > 0 )     return A[left];
         else      return 0 ;

    Center = (Left + Right) / 2 ;
    MaxLeftSum = MaxSubSum(A,Left,Center);
    MaxRightSum = MaxSubSum(A,Center + 1 ,Right);

    MaxLeftBorderSum = 0 ;
    LeftBorderSum = 0 ;
     for (i = Center;i >= Left;i -- )
     {
        LeftBorderSum += A[i];
         if (LeftBorderSum > MaxLeftBorderSum)    MaxLeftBorderSum = LeftBorderSum;
    }

    MaxRightBorderSum = 0 ;
    RightBorderSum = 0 ;
     for (i = Center + 1 ;i <= Right;i ++ )
     {
        RightBorderSum += A[i];
         if (RightBorderSum > MaxRightBorderSum)    MaxRightBorderSum = RightBorderSum;
    }

     return Max3(MaxLeftSum,MaxRightSum,MaxLeftBorderSum + MaxRightBorderSum);
}

int MaxSubsequenceSum( const int  A[],int N)
{
     return MaxSubSum(A, 0 ,N - 1 );
}
旉��O(N)�Q�算法四

intMaxSubsequenceSum( const int A[], int N)
{
     int ThisSum,MaxSum,i;

    ThisSum = MaxSum = 0 ;
     for (i = 0 ;i < N;i ++ )
     {
        ThisSum += A[i];
         if (ThisSum > MaxSum)
            MaxSum = ThisSum;
         else
            ThisSum = 0 ;
    }

     return MaxSum;
}

参考《数据结构与��法分析�?/font>

Dain 2007-02-07 10:52 发表评论

Dain — Mon, 25 Dec 2006 02:17:00 GMT

首先�Q�给出算法的思�\
设R={r₁,r₂,...,r_n}是要�q�行排列的n个元素，R_i=R-{r_i}�?br />集合X中元素的全排列记为permutation(X)�Q?ri)permutation(X)表示在全排列permutation(X)的每一个排列前加上前缀r_i得到的排列�?br />R的全排列可归�U�_��义如下：(x��)
当n=1�Ӟ��permutation(R)={r},r是集合R中唯一的元素；
当n>1�Ӟ��permutation(R)�?r₁)permutation(R₁)�Q?r₂)permutation(R₂)�Q�……，(r_n)permutation(R_n)构成�?br />
此算法要求待排列的数据是互异的，因�ؓ(f��)该算法不能检��同�U�排列是否已�l�输出，如：(x��)
1, 1, 2
那么�Q�全排列期望输出是：(x��)
1, 1, 2
1, 2, 1
2, 1, 1
但是该算法的输出�Q?br /> 1, 1, 2
1, 2, 1
2, 1, 1
1, 1, 2
1, 2, 1
2, 1, 1
�q�是该算法的�~�点�Q�也限制�?ji��n)它的适用范围�?br />
�E�序描述如下�Q?/font>

#include < iostream >
#include < algorithm >

using namespace std;

// 递归产生R[k:n]的所有的排列�Q?strong>元素是互异的
template < class Type >
void permutation(Type * R, int k, int n)
{
     if (k == n)
    {
         for ( int i = 0 ;i < n; ++ i)
            cout << R[i] << " \t " ;
        cout << endl;
    }
     else
         for ( int i = k;i < n; ++ i)
        {
            swap(R[k],R[i]);
            permutation(R,k + 1 ,n);
            swap(R[k],R[i]);
        }
}

�q�有一�U�很��单的�Ҏ(gu��)��Q��用GP中的�Ҏ(gu��)��

该算法是STL中的范型��法�Q�当然效果是很好的，不会(x��)出现上面的算法的情况�?br />
�E�序描述如下�Q?/font>

// 使用泛型��法next_permutation()
#include < iostream >
#include < vector >
#include < algorithm >

using namespace std;

// 产生R[k:n]的所有的排列
template < class Type >

void pernutation(Type * R, int k, int n)
{
vector < Type > myVec;
int i,size = n - k;
for (i = k;i < n;i ++ )
  myVec.push_back(R[i]);
// 使用next_permutation()函数必须是有序的数据
sort(myVec.begin(),myVec.end());

do
{
   for (i = 0 ;i < size;i ++ )
   cout << myVec[i] << " \t " ;
  cout << endl;
}
while (next_permutation(myVec.begin(),myVec.end()));
}

注：(x��)�q�里的待全排的数据是存在数组或者向量中的�?/p>

Dain 2006-12-25 10:17 发表评论

��法求解(�q�没有�h做出来，�?

Dain — Thu, 21 Dec 2006 02:58:00 GMT
     摘要: Given an int m and a vector a containing n numbers, return the result of the following expression:   阅读全文

Dain 2006-12-21 10:58 发表评论

string和int间的�c�d��转换

Dain — Sun, 10 Dec 2006 05:05:00 GMT
int 转换 string

string int2str( int num)
{
    if (num == 0 )
       return " 0 " ;

    string str = "" ;
    int num_ = num > 0 ? num : - 1 * num;

    while (num_)
    {
      str = ( char )(num_ % 10 + 48 ) + str;
      num_ /= 10 ;
    }

    if (num < 0 )
      str = " - " + str;

    return str;
}
string 转换 int

int str2int( string str)
{
    int i,len = str.size(),num = 0 ;

    i = 0 ;
    if (str[ 0 ] == ' - ' )
      i = 1 ;

    while (i < len)
    {
      num = num * 10 + ( int )(str[i] - ' 0 ' );
      i ++ ;
    }

    if (str[ 0 ] == ' - ' )
      num *= - 1 ;

    return num;
}

string和int间是可以自己写函数实现类型�{换的�Q�但是string和double��׃��太方便了(ji��n)�Q�stirng转换double可以自定义函数实玎ͼ�但是�Q�double转换string��需要��用库函数�?ji��n)，如sprintf

Dain 2006-12-10 13:05 发表评论

国内高清久久久久久,国内精品伊人久久久久影院对白,漂亮人妻被黑人久久精品

列出所�?位数�Q�它的前n位能被n整除

最大的子序列和问题

���法求解(�q�没有�h做出来，�?

string和int间的�c�d��转换

��法求解(�q�没有�h做出来，�?