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tate

Dain — Thu, 18 Sep 2008 11:58:00 GMT

Dain 2008-09-18 19:58 发表评论

Getting the Minimum and Maximum Values for a Numeric Type

Dain — Tue, 29 May 2007 02:38:00 GMT

Getting numeric limits

#include <iostream>
#include <limits>

using namespace std;

template<typename T>
void showMinMax() {
   cout << "min: " << numeric_limits<T>::min() << endl;
   cout << "max: " << numeric_limits<T>::max() << endl;
   cout << endl;
}

int main() {
   cout << "short:" << endl;
   showMinMax<short>();
   cout << "int:" << endl;
   showMinMax<int>();
   cout << "long:" << endl;
   showMinMax<long>();
   cout << "long long:" << endl;
   showMinMax<long long>();
   cout << "float:" << endl;
   showMinMax<float>();
   cout << "double:" << endl;
   showMinMax<double>();
   cout << "long double:" << endl;
   showMinMax<long double>();
   cout << "unsigned short:" << endl;
   showMinMax<unsigned short>();
   cout << "unsigned int:" << endl;
   showMinMax<unsigned int>();
   cout << "unsigned long:" << endl;
   showMinMax<unsigned long>();
   cout << "unsigned long long:" << endl;
   showMinMax<unsigned long long>();
}

Dain 2007-05-29 10:38 发表评论

3017

Dain — Fri, 25 May 2007 02:06:00 GMT

#include <stdio.h>
#include <stdlib.h>
#include <vector>

using namespace std;

struct Node {
    int i,j;
    int value;
};

long num[100000];
vector<Node> matrix;

int main() {
    long n;
    long long m;
    scanf("%ld %lld",&n,&m);

    long i,j;
    for(i = 0;i < n;++i) {
        scanf("%ld",&num[i]);
    }

    for(i = 0;i < n;++i) {
        if(num[i] > m) {
            break;
        }
    }

    if(i < n) {
        printf("-1\n");

        return 0;
    }

    long long res = -1;
    long long sum;
    long max,min = 0;
    for(i = 0;i < n;++i) {
        if(i > 0) {
            min = 1000000;
            for(j = 0;j < matrix.size();++j) {
                if(matrix[j].j == i - 1 && matrix[j].value < min) {
                    min = matrix[j].value;
                }
            }
        }
        else {
            min = 0;
        }

        sum = 0;
        Node node;
        max = -1;
        for(j = i;j < n;++j) {
            sum += num[j];
            if(sum <= m) {
                if(max < num[j]) {
                    max = num[j];
                }
                node.i = i;
                node.j = j;
                node.value = max + min;
                matrix.push_back(node);
                if(j == n - 1) {
                    if(res != -1) {
                        if(node.value < res) {
                            res = node.value;
                        }
                    }
                    else {
                        res = node.value;
                    }
                }
            }
            else {
                break;
            }
        }
    }

    printf("%lld\n",res);

    return 0;
}

Dain 2007-05-25 10:06 发表评论

3017

Dain — Fri, 25 May 2007 02:06:00 GMT

Dain 2007-05-25 10:06 发表评论

Dain — Thu, 24 May 2007 05:56:00 GMT

最�q�，��L��犯非�怽��U�的错误
看题不仔�l?br>��j误写成k�Q�而且怪的是，��试的例子都通过了，后来通过debug才找��C��q�个很低�U�的错误

真是气�h�?br>
不要再犯�?/span>

Dain 2007-05-24 13:56 发表评论

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

�~�写递归四条基本法则

Dain — Wed, 31 Jan 2007 13:03:00 GMT

基准情�Ş。必��要有某些基准情形，它无��递归��p��解出�Q�也��是要有退出递归的条件�?/font>

不断推进。对于那些需要递归求解的情形，每一�ơ递归调用都必��要使求解状冉|��接近基准情�Ş的方向推�q��?/font>

设计法则。假设所有的递归调用都能�q�行�?/font>

合成效益。求解一个问题的同一个实例时�Q�切勿在不同的递归调用中做重复性的工作�?/font>

Dain 2007-01-31 21:03 发表评论

��M��计划

Dain — Wed, 31 Jan 2007 12:07:00 GMT
遇到了好多不能解决的问题后，觉得应该重新读读书了
最�q�买了几本书
《More Effective CPP�?br />《Effective STL�?br />《�ƈ行程序设计�?br />《数据结构与��法分析——C语言描述�?/font>

Dain 2007-01-31 20:07 发表评论

引用和指针参数的关系

Dain — Fri, 19 Jan 2007 01:56:00 GMT
两种参数都允许函��C��改实参指向的对象�Q�都允许有效地向函数传递大�c�d��对象。所以怎么样决定把函数参数声明成引用还是指针呢�Q?br />引用必须被初始化为指向一个对象，一旦初始化了，它就不能再指向其他对象。指针可以指向一�p�d��不同的对象也可以什么都不指向�?br />因�ؓ指针可能指向一个对象或没有��M��对象�Q�所以函数在��定指针实际指向一个有效的对象之前不能安全解引用一个指针。如�Q?br />
class X;
void fun(X * x)
{
   // 在解引用指针之前��信它非0
   if (x != 0 )
     // 解引用指�?/span>
}
而，对于引用参数�Q�函��C��需要保证它指向一个对象。引用必��L��向一个对象，不希望向指针那样�q�行解引用。如�Q?br />
class Type;
void op(const Type &t1,const Type &t2);

int main()
{
  Type obj1;
  // 讄��obj1为某个�?br />
  // 错误�Q�引用参数的实参不能�?
  op(obj1,0);

  //
  return 0;
}
如果一个参数可能在函数中指向不同的对象�Q�或者这个参数可能不指向��M��对象�Q�则必须使用指针参数�?br />引用参数的一个重要用法，它允许有效地实现重蝲操作�W�的同时�Q�还能保证用法的直观性。可以参考《C++ Primer�?br />
ps 发现�?87��늚��W�二个程序例子是错的

Dain 2007-01-19 09:56 发表评论