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STL 容器Deque

patriking — Fri, 21 Jan 2011 14:25:00 GMT

template <class T, template <class U> class Allocator = allocator>
class deque {
public:
// typedefs:
    typedef iterator;
    typedef const_iterator;
    typedef Allocator<T>::pointer pointer;
    typedef Allocator<T>::reference reference;
    typedef Allocator<T>::const_reference const_reference;
    typedef size_type;
    typedef difference_type;
    typedef T value_type;
    typedef reverse_iterator;
    typedef const_reverse_iterator;
// allocation/deallocation:
    deque();
    deque(size_type n, const T& value = T());
    deque(const deque<T, Allocator>& x);
    template <class InputIterator>
    deque(InputIterator first, InputIterator last);
    ~deque();
    deque<T, Allocator>& operator=(const deque<T, Allocator>& x);
    void swap(deque<T, Allocator>& x);
// accessors:
    iterator begin();
    const_iterator begin() const;
    iterator end();
    const_iterator end() const;
    reverse_iterator rbegin();
    const_reverse_iterator rbegin();
    reverse_iterator rend();
    const_reverse_iterator rend();
    size_type size() const;
    size_type max_size() const;
    bool empty() const;
    reference operator[](size_type n);
    const_reference operator[](size_type n) const;
    reference front();
    const_reference front() const;
    reference back();
    const_reference back() const;
// insert/erase:
    void push_front(const T& x);
    void push_back(const T& x);
    iterator insert(iterator position, const T& x = T());
    void insert (iterator position, size_type n, const T& x);
    template <class InputIterator>
    void insert (iterator position, InputIterator first, InputIterator last);
    void pop_front();
    void pop_back();
    void erase(iterator position);
    void erase(iterator first, iterator last);
};

patriking 2011-01-21 22:25 发表评论

STL 容器List

patriking — Fri, 21 Jan 2011 14:15:00 GMT

template <class T, template <class U>
class Allocator = allocator> class list
{
public:
// typedefs:
    typedef iterator
    typedef const_iterator
    typedef Allocator<T>::pointer pointer
    typedef Allocator<T>::reference reference
    typedef Allocator<T>::const_reference const_reference
    typedef size_type
    typedef difference_type
    typedef T value_type
    typedef reverse_iterator
    typedef const_reverse_iterator;
// allocation/deallocation:
    list()
    list(size_type n, const T& value = T())
    template <class InputIterator>
    list(InputIterator first, InputIterator last)
    list(const list<T, Allocator>& x)
    ~list()
    list<T, Allocator>& operator=(const list<T, Allocator>& x)
    void swap(list<T, Allocator>& x);
// accessors:
    iterator begin()
    const_iterator begin() const
    iterator end()
    const_iterator end() const
    reverse_iterator rbegin()
    const_reverse_iterator rbegin();
    reverse_iterator rend();
    const_reverse_iterator rend();
    bool empty() const;
    size_type size() const;
    size_type max_size() const;
    reference front();
    const_reference front() const;
    reference back();
    const_reference back() const;
// insert/erase:
    void push_front(const T& x);
    void push_back(const T& x);
    iterator insert(iterator position, const T& x = T());
    void insert(iterator position, size_type n, const T& x);
    template <class InputIterator>
    void insert(iterator position, InputIterator first, InputIterator last);
    void pop_front();
    void pop_back();
    void erase(iterator position);
    void erase(iterator first, iterator last);
// special mutative operations on list:
    void splice(iterator position, list<T, Allocator>& x);
    void splice(iterator position, list<T, Allocator>& x, iterator i);
    void splice(iterator position, list<T, Allocator>& x,
    iterator first, iterator last);
    void remove(const T& value);
    template <class Predicate> void remove_if(Predicate pred);
    void unique();
    template <class BinaryPredicate> void unique(BinaryPredicate binary_pr
    void merge(list<T, Allocator>& x);
    template <class Compare> void merge(list<T, Allocator>& x, Compare com
    void reverse();
    void sort();
    template <class Compare> void sort(Compare comp);
};

patriking 2011-01-21 22:15 发表评论

最��割 ZOJ@2788

patriking — Thu, 20 Jan 2011 07:52:00 GMT

最大流最��割定理介绍�Q�把一个流�|�络的顶炚w��划分成两个集合S和T�Q��得源点s ∈S且汇点t ∈T�Q�割(S,T)的容量C(S,T) = �?Cuv, 其中u∈S且v∈T�?br>从直观上看，截集(S,T)是从源点s到汇点t的必�l�之路，如果该�\堵塞则流从s无法到达t。于是我们可以得�?span style="color: red;">最大流最��割定理�Q��Q意一个流�|�络的最大流量等于该�|�络的最��的割的定w��?br> 最��割问题�Q?a >ZOJ@2788
题意�Q�在一栋楼中有若干戉K��Q�一个房间有若干扇门通往别的戉K��Q�现在房间中有一个secure room。由于有一些敌人已�l�潜入某一些房��_��现在问关闭最��的一些门�Q��得敌��Z��到达不了secure room。对于一扇门�q�通A-B�Q�若控制面板CP在A方，则��L��可以从A->B�Q�但是若关上门，则B不能到达A�?br> 解法�Q�最大流。首先通过floyd��法判断是否敌�h��L��可以到达secure room�Q�如果能直接输出�l�果�Q�如果不能，则通过最大流��法求解卛_��。对于一扇门A-B�Q�CP在A方则从A�q�一条有向边到B�Q�容量�ؓ无穷大，从B�q�一条边到A�Q�容量�ؓ1。两个房间之间可能有多扇门，边的定w��累加卛_��。增加一个源点，从源点连边到有敌人的戉K��Q�容量�ؓ1。对图求最大流即�ؓ�l�果�?br>代码如下�Q?br>

// 2389732      2011-01-20 15:06:16        Accepted      2788      C++      0      184      *******
#include<cstdio>
#include<algorithm>
#include<cstring>
#define MAXN 22
#define inf 1000000
int map[MAXN][MAXN];
int flow[MAXN][MAXN];
int max_flow(int n,int mat[][MAXN],int source,int sink,int flow[][MAXN]){
    int pre[MAXN],que[MAXN],p,q,t,i,j;
    int d[MAXN];
    int tmp;
    if (source==sink) return inf;
    for (i=0;i<n;i++)
        for    (j=0;j<n;flow[i][j++]=0);
    for (;;){
        for (i=0;i<n;pre[i++]=0);
        pre[t=source]=source+1,d[t]=inf;
        for (p=q=0;p<=q&&!pre[sink];t=que[p++])
               for (i=0;i<n;i++)
                if (!pre[i]&&(tmp=mat[t][i]-flow[t][i]))
                     pre[que[q++]=i]=t+1,d[i]=d[t]<tmp?d[t]:tmp;
                else if (!pre[i]&&(tmp=flow[i][t]))
                 pre[que[q++]=i]=-t-1,d[i]=d[t]<tmp?d[t]:tmp;
        if (!pre[sink]) break;
        for (i=sink;i!=source;)
               if (pre[i]>0)
                flow[pre[i]-1][i]+=d[sink],i=pre[i]-1;
               else
                flow[i][-pre[i]-1]-=d[sink],i=-pre[i]-1;
    }
    tmp = 0;
    for (i=0;i<n;tmp+=flow[source][i++]);
    return tmp;
}
int main()
{
    int T, n, m, k, y;
    char str[3];
    int d[MAXN];
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&m,&n);
        n++;
        memset(map,0,sizeof(map));
        memset(flow,0,sizeof(flow));
        for(int i = 1; i <= m; i++){
            scanf("%s %d",str, &k);
            d[i] = (strlen(str) == 1);
            if(d[i])map[0][i] = inf;
            for(int j = 0; j < k; j++){
                scanf("%d",&y);
                y++;
                flow[i][y] = 1;
                map[i][y] += inf;
                map[y][i] += 1;
            }
        }
        // Checking always exists a path to secure room. Output "PANIC ROOM BREACH"!
        for(k = 1; k <= m; k++)
            for(int i = 1; i <= m; i++)
                for(int j = 1; j <= m; j++)
                    if(flow[i][k]&&flow[k][j])
                        flow[i][j] = 1;
        for(k = 1; k <= m && !(d[k] && flow[k][n]); k++);
        if(k <= m)
            printf("PANIC ROOM BREACH\n");
        else
            printf("%d\n", max_flow(m+1,map,0,n,flow));
    }
    return 0;
}

patriking 2011-01-20 15:52 发表评论

patriking — Wed, 19 Jan 2011 11:57:00 GMT

int abs(int i) �q�回整型参数i的绝对�?br>double cabs(struct complex znum) �q�回复数znum的绝对�?br>double fabs(double x) �q�回双精度参数x的绝对�?br>long labs(long n) �q�回长整型参数n的绝对�?br>double exp(double x) �q�回指数函数e^x的�?br>double frexp(double value,int *eptr) �q�回value=x*2^n中x的�?0.5<=x<1),n存贮在eptr�?br>double ldexp(double value,int exp); �q�回value*2^exp的�?br>double log(double X) �q�回logeX的�?以e为底的对�?
double log10(double x) �q�回log10x的�?�?0为底的对�?br>double pow(double x,double y) �q�回x^y的�?br>double pow10(int p) �q�回10^p的�?br>double sqrt(double x) �q�回a,且a满��a*a=x,a>=0
double acos(double x) �q�回x的反余��ucos^-1(x)�?x为弧�?br>double asin(double x) �q�回x的反正��usin^-1(x)�?x为弧�?br>double atan(double x) �q�回x的反正切tan^-1(x)�?x为弧�?br>double atan2(double y,double x) �q�回y/x的反正切tan^-1(x)�?y的x为弧�?br>double cos(double x) �q�回x的余弦cos(x)�?x为弧�?br>double sin(double x) �q�回x的正弦sin(x)�?x为弧�?br>double tan(double x) �q�回x的正切tan(x)�?x为弧�?br>double cosh(double x) �q�回x的双曲余弦cosh(x)�?x为弧�?br>double sinh(double x) �q�回x的双曲正弦sinh(x)�?x为弧�?br>double tanh(double x) �q�回x的双曲正切tanh(x)�?x为弧�?br>double hypot(double x,double y) �q�回直角三角形斜边的长度(z),x和y为直角边的长�?z^2=x^2+y^2
double ceil(double x) �q�回不小于x的最��整�?br>double floor(double x) �q�回不大于x的最大整�?br>void srand(unsigned seed) 初始化随机数发生�?br>int rand() 产生一个随机数�q�返回这个数
double poly(double x,int n,double c[])从参��C�生一个多��式
double modf(double value,double *iptr)��双�_�ֺ�数value分解成整数部分和��数部分�Q�iptr为整数部分，�q�回��数部分
double fmod(double x,double y) �q�回x/y的余�?br>double atof(char *nptr) ��字�W�串nptr转换成��Q�Ҏ��q�返回这个��Q�Ҏ��
double atoi(char *nptr) ��字�W�串nptr转换成整数�ƈ�q�回�q�个整数
double atol(char *nptr) ��字�W�串nptr转换成长整数�q�返回这个整�?br>char *ecvt(double value,int ndigit,int *decpt,int *sign)��Q�Ҏ��value转换成字�W�串�q�返回该字符�?br>char *fcvt(double value,int ndigit,int *decpt,int *sign)��Q�Ҏ��value转换成字�W�串�q�返回该字符�?br>char *gcvt(double value,int ndigit,char *buf)��数value转换成字�W�串�q�存于buf�?�q�返回buf的指�?br>char *ultoa(unsigned long value,char *string,int radix)��无�W�号整型数value转换成字�W�串�q�返回该字符�?radix��{换时所用基�?br>char *ltoa(long value,char *string,int radix)��长整型数value转换成字�W�串�q�返回该字符�?radix��{换时所用基�?br>char *itoa(int value,char *string,int radix)��整数value转换成字�W�串存入string,radix��{换时所用基�?br>double atof(char *nptr) ��字�W�串nptr转换成双�_�ֺ��?�q�返回这个数,错误�q�回0
int atoi(char *nptr) ��字�W�串nptr转换成整型数, �q�返回这个数,错误�q�回0
long atol(char *nptr) ��字�W�串nptr转换成长整型�?�q�返回这个数,错误�q�回0
double strtod(char *str,char **endptr)��字�W�串str转换成双�_�ֺ��?�q�返回这个数,
long strtol(char *str,char **endptr,int base)��字�W�串str转换成长整型�?�q�返回这个数,
int matherr(struct exception *e)用户修改数学错误�q�回信息函数
double _matherr(_mexcep why,char *fun,double *arg1p,
double *arg2p,double retval)用户修改数学错误�q�回信息函数
unsigned int _clear87() 清除��点状态字�q�返回原来的��点状�?br>void _fpreset()重新初��化��Q�Ҏ��学程序包
unsigned int _status87() �q�回��点状态字

patriking 2011-01-19 19:57 发表评论

STL之Set operation on sorted structures

patriking — Tue, 18 Jan 2011 02:17:00 GMT

Set operations on sorted structures 为STL中对两个有序序列的操作，包括包含、合�q�、求交集、求差、求对称差�?br>包含�Q�如果[first2,last2)包含在[first1,last1)中，�q�回true,否则�q�回false�?br>bool includes(InputIterator1 first1, InputIterator1 last1,InputIterator2 first2, InputIterator2 last2)
bool includes(InputIterator1 first1, InputIterator1 last1,InputIterator2 first2, InputIterator2 last2, Compare comp)

合�ƈ�Q�合�q�[first1,last1)与[first2,last2)两个集合�?br>OutputIterator set_union(InputIterator1 first1, InputIterator1 last1, InputIterator2 first2, InputIterator2 last2, OutputIterator result)
OutputIterator set_union(InputIterator1 first1, InputIterator1 last1, InputIterator2 first2, InputIterator2 last2, OutputIterator result, Compare comp)

求交集：求[first1,last1)与[first2,last2)两个集合的交集�?
OutputIterator set_intersection(InputIterator1 first1, InputIterator1 last1, InputIterator2 first2, InputIterator2 last2, OutputIterator result)
OutputIterator set_intersection(InputIterator1 first1, InputIterator1 last1, InputIterator2 first2, InputIterator2 last2, OutputIterator result, Compare comp)

求差�Q�求在[first1,last1)中，而不在[first2,last2)的元素集合�?
OutputIterator set_difference(InputIterator1 first1, InputIterator1 last1, InputIterator2 first2, InputIterator2 last2, OutputIterator result)
OutputIterator set_difference(InputIterator1 first1, InputIterator1 last1, InputIterator2 first2, InputIterator2 last2, OutputIterator result, Compare comp)

求对�U�差�Q�两个集合之和集减去两个集合之交集�?
OutputIterator set_symmetric difference(InputIterator1 first1, InputIterator1 last1, InputIterator2 first2, InputIterator2 last2, OutputIterator result)
OutputIterator set_symmetric difference(InputIterator1 first1, InputIterator1 last1, InputIterator2 first2, InputIterator2 last2, OutputIterator result, Compare comp)

#include<iostream>
#include<algorithm>
#include<vector>
#include<iterator>
using namespace std;
void init(vector<int>& v, int a, int b)
{
    v.clear();
    for(int i = a; i <= b; i++){
        v.push_back(i);
    }
}
int main()
{
    vector<int>v1, v2;
    vector<int>::iterator itr;
    cout << "Includes operator:" << endl;
    init(v1,1,5);
    init(v2,1,8);
    cout << "Vector v1: ";
    copy(v1.begin(),v1.end(),ostream_iterator<int>(cout," "));
    cout << endl;
    cout << "Vector v2: ";
    copy(v2.begin(),v2.end(),ostream_iterator<int>(cout," "));
    cout << endl;
    cout << "v2 is in v1 ? " << (includes(v1.begin(),v1.end(),v2.begin(),v2.end()) ? "yes" : "no") << endl;
    cout << "v1 is in v2 ? " << (includes(v2.begin(),v2.end(),v1.begin(),v1.end()) ? "yes" : "no") << endl;

    cout << endl;
    init(v1,1,5);
    init(v2,3,8);
    cout << "Vector v1: ";
    copy(v1.begin(),v1.end(),ostream_iterator<int>(cout," "));
    cout << endl;
    cout << "Vector v2: ";
    copy(v2.begin(),v2.end(),ostream_iterator<int>(cout," "));
    cout << endl;
    cout << "set_union operator on v1,v2: ";
    set_union(v1.begin(),v1.end(),v2.begin(),v2.end(),ostream_iterator<int>(cout," "));
    cout << endl << endl;

    init(v1,1,5);
    init(v2,3,8);
    cout << "Vector v1: ";
    copy(v1.begin(),v1.end(),ostream_iterator<int>(cout," "));
    cout << endl;
    cout << "Vector v2: ";
    copy(v2.begin(),v2.end(),ostream_iterator<int>(cout," "));
    cout << endl;
    cout << "set_intersecion operator on v1,v2: ";
    set_intersection(v1.begin(),v1.end(),v2.begin(),v2.end(),ostream_iterator<int>(cout," "));
    cout << endl << endl;

    init(v1,1,5);
    init(v2,3,8);
    cout << "Vector v1: ";
    copy(v1.begin(),v1.end(),ostream_iterator<int>(cout," "));
    cout << endl;
    cout << "Vector v2: ";
    copy(v2.begin(),v2.end(),ostream_iterator<int>(cout," "));
    cout << endl;
    cout << "set_difference operator, v1 to v2: ";
    set_difference(v1.begin(),v1.end(),v2.begin(),v2.end(),ostream_iterator<int>(cout," "));
    cout << endl;
    cout << "set_difference operator, v2 to v1: ";
    set_difference(v2.begin(),v2.end(),v1.begin(),v1.end(),ostream_iterator<int>(cout," "));
    cout << endl << endl;

    init(v1,1,5);
    init(v2,3,8);
    cout << "Vector v1: ";
    copy(v1.begin(),v1.end(),ostream_iterator<int>(cout," "));
    cout << endl;
    cout << "Vector v2: ";
    copy(v2.begin(),v2.end(),ostream_iterator<int>(cout," "));
    cout << endl;
    cout << "set_symmetric_difference operator, v1 to v2: ";
    set_symmetric_difference(v1.begin(),v1.end(),v2.begin(),v2.end(),ostream_iterator<int>(cout," "));
    cout << endl;
    cout << "set_symmetric_difference operator, v2 to v1: ";
    set_symmetric_difference(v2.begin(),v2.end(),v1.begin(),v1.end(),ostream_iterator<int>(cout," "));
    cout << endl << endl;
    return 0;
}

Includes operator:
Vector v1: 1 2 3 4 5
Vector v2: 1 2 3 4 5 6 7 8
v2 is in v1 ? no
v1 is in v2 ? yes

Vector v1: 1 2 3 4 5
Vector v2: 3 4 5 6 7 8
set_union operator on v1,v2: 1 2 3 4 5 6 7 8

Vector v1: 1 2 3 4 5
Vector v2: 3 4 5 6 7 8
set_intersecion operator on v1,v2: 3 4 5

Vector v1: 1 2 3 4 5
Vector v2: 3 4 5 6 7 8
set_difference operator, v1 to v2: 1 2
set_difference operator, v2 to v1: 6 7 8

Vector v1: 1 2 3 4 5
Vector v2: 3 4 5 6 7 8
set_symmetric_difference operator, v1 to v2: 1 2 6 7 8
set_symmetric_difference operator, v2 to v1: 1 2 6 7 8

patriking 2011-01-18 10:17 发表评论

STL之Merge

patriking — Mon, 17 Jan 2011 13:35:00 GMT

STL中合�q�算法merge介绍�Q�merge与inplace_merge�?br>merge合�ƈ两个有序的sequence�Q�默认�ؓ<比较�Q�可自定义比较函数comp�Q�结果保存在[result, result + (last1 - first1) + (last2 - first2)) 中。函数定义如下：
OutputIterator merge (InputIterator1 first1, InputIterator1 last1, InputIterator2 first2, InputIterator2 last2, OutputIterator result)
OutputIterator merge (InputIterator1 first1, InputIterator1 last1, InputIterator2 first2, InputIterator2 last2, OutputIterator result, Compare comp)

inplace_merge合�ƈ一个中前后两段分别有序的序列，即[first,middle)与[middle,last)分别有序�Q�合�q�这两个子序列，�q�把�l�果保存在[first,last�Q�中�?br>inplace_merge (BidirectionalIterator first, BidirectionalIterator middle, BidirectionalIterator last);
inplace_merge (BidirectionalIterator first, BidirectionalIterator middle, BidirectionalIterator last, Compare comp);

#include<string>
#include<algorithm>
#include<iostream>
#include<iterator>
#include<vector>
using namespace std;
int main()
{
    vector<int>v1;
    vector<int>v2;
    vector<int>v3;
    vector<int>::iterator vecitr;
    for(int i = 1; i <= 10; i++){
        v1.push_back(i%7);
        v2.push_back(i%4);
    }
    sort(v1.begin(),v1.end());
    cout << "Fisrt vecter: ";
    copy(v1.begin(),v1.end(),ostream_iterator<int>(cout," "));
    cout << endl << endl;
    sort(v2.begin(),v2.end());
    cout << "Second vecter: ";
    copy(v2.begin(),v2.end(),ostream_iterator<int>(cout," "));
    cout << endl << endl;
    cout << "Merge this two vecter: ";
    merge(v1.begin(),v1.end(),v2.begin(),v2.end(),ostream_iterator<int>(cout," "));
    cout << endl << endl;
    for(int i = 1; i <= 10; i++){
        v3.push_back(i%7);
    }
    sort(v3.begin(),v3.end());
    int x = v3.size();
    for(int i = 1; i <= 10; i++){
        v3.push_back(i%4);
    }
    vecitr = v3.begin();
    advance(vecitr,x);
    sort(vecitr,v3.end());
    cout << "Third vecter: ";
    copy(v3.begin(),v3.end(),ostream_iterator<int>(cout," "));
    cout << endl << endl;
    cout << "inplace_Merge third vecter: ";
    inplace_merge(v3.begin(),vecitr,v3.end());
    copy(v3.begin(),v3.end(),ostream_iterator<int>(cout," "));
    cout << endl << endl;
    return 0;
}

Fisrt vecter: 0 1 1 2 2 3 3 4 5 6

Second vecter: 0 0 1 1 1 2 2 2 3 3

Merge this two vecter: 0 0 0 1 1 1 1 1 2 2 2 2 2 3 3 3 3 4 5 6

Third vecter: 0 1 1 2 2 3 3 4 5 6 0 0 1 1 1 2 2 2 3 3

inplace_Merge third vecter: 0 0 0 1 1 1 1 1 2 2 2 2 2 3 3 3 3 4 5 6

patriking 2011-01-17 21:35 发表评论

STL之Binary search

patriking — Sun, 16 Jan 2011 14:40:00 GMT

STL中对于有序序列（vector,list�{�）提供了相当相当强大的二分搜烦Binary search��法。对于可以随��问容器（如vector�{�）�Q�binary search负蝲度�ؓ�Ҏ��U�别�Q�LogN)�Q�对于非随机讉K��容器�Q�如list�Q�，则算法复杂度为线性。现在简要介�l�一下几�U�常用的binary search��法�Q?br>
ForwardIterator lower_bound (ForwardIterator first,ForwardIterator last, const T& value)
ForwardIterator lower_bound (ForwardIterator first,ForwardIterator last, const T& value, Compare comp)
//查找最�q�的一个iterator i�Q?使得[first,i)里面的�Q意一个iterator j�Q�有 *j < value or comp(*j,value)==true.

ForwardIterator upper_bound (ForwardIterator first,ForwardIterator last, const T& value)
ForwardIterator upper_bound (ForwardIterator first,ForwardIterator last, const T& value, Compare comp)
//查找最�q�的一个iterator i�Q?使得[first,i)里面的�Q意一个iterator j�Q�有 !(*j < value) or comp(*j,value)==false.

pair equal_range(ForwardIterator first, ForwardIterator last, const T& value)
pair equal_range(ForwardIterator first, ForwardIterator last, const T& value, Compare comp)
//查找最大的subrange[i,j)�Q��得�Q意一个iterator in [i,j)满��!(*k
bool binary_search (ForwardIterator first, ForwardIterator last, const T& value)
bool binary_search (ForwardIterator first, ForwardIterator last, const T& value, Compare comp)
// 查找是否在[first,last)中存在iterator i,满�� !(*i// 存在则返回true,否则�q�回false.

#include<algorithm>
#include<iostream>
#include<vector>
#include<iterator>
using namespace std;
int main()
{
    vector<int>v;
    vector<int>::iterator itr;
    pair< vector<int>::iterator, vector<int>::iterator >vecpair;

    for(int i = 1; i <= 20; i++){
        v.push_back(i%6);
    }
    sort(v.begin(),v.end());
    cout << "array: " << endl << "    ";
    copy(v.begin(),v.end(),ostream_iterator<int>(cout," "));
    cout << endl;

    //lower_bound
    cout << "lower_bound function, value = 3:" << endl;
    itr = lower_bound(v.begin(),v.end(),3);
    cout << "    [first, itr) = ";
    copy(v.begin(),itr,ostream_iterator<int>(cout," "));
    cout << endl;
    cout << "    [itr, last) = ";
    copy(itr,v.end(),ostream_iterator<int>(cout," "));
    cout << endl << endl;

    // upper_bound
    cout << "upper_bound function, value = 3:" << endl;
    itr = upper_bound(v.begin(),v.end(),3);
    cout << "    [first, itr) = ";
    copy(v.begin(),itr,ostream_iterator<int>(cout," "));
    cout << endl;
    cout << "    [itr, last) = ";
    copy(itr,v.end(),ostream_iterator<int>(cout," "));
    cout << endl << endl;

    // equal_range
    cout << "equal_range function, value = 3:" << endl;
    vecpair = equal_range(v.begin(),v.end(),3);

    cout << "    [vecpair->first, vecpair->second) = ";
    copy(vecpair.first, vecpair.second, ostream_iterator<int>(cout," "));
    cout << endl << endl;

    //binary_search, value = 3
    cout << "binary_search function, value = 3:" << endl;
    cout << "3 is " << (binary_search(v.begin(),v.end(),3) ? "":"not ") << "in array." << endl;
    cout << endl;

    //binary_search, value = 6
    cout << "binary_search function, value = 6:" << endl;
    cout << "6 is " << (binary_search(v.begin(),v.end(),6) ? "":"not ") << "in array." << endl;
    cout << endl;

    return 0;
}

array:
    0 0 0 1 1 1 1 2 2 2 2 3 3 3 4 4 4 5 5 5
lower_bound function, value = 3:
    [first, itr) = 0 0 0 1 1 1 1 2 2 2 2
    [itr, last) = 3 3 3 4 4 4 5 5 5

upper_bound function, value = 3:
    [first, itr) = 0 0 0 1 1 1 1 2 2 2 2 3 3 3
    [itr, last) = 4 4 4 5 5 5

equal_range function, value = 3:
    [vecpair->first, vecpair->second) = 3 3 3

binary_search function, value = 3:
3 is in array.

binary_search function, value = 6:
6 is not in array.

patriking 2011-01-16 22:40 发表评论

sscanf and scanf函数%[]用法

patriking — Sat, 15 Jan 2011 11:54:00 GMT

scanf函数中的%[]用法�Q?br>    %[]表示要读入一个字�W�集�? 如果[后面�W�一个字�W�是”^”�Q�则表示反意思�?br>    []内的字符串可以是1或更多字�W�组成。空字符集（%[]�Q�是�q�反规定的，可导致不可预知的�l�果�Q�同�?[^]也是�q�反规定的�?br>例如: %[a-z]��d��在a-z之间的字�W�串�Q�如果不在此之前则停止，�?br>          char s[]="hello, my friend” ; // 注意: ,逗号在不 a-z之间
          sscanf( s, “%[a-z]”, string ) ; // string=hello
      %[^a-z] ��d��不在a-z之间的字�W�串�Q�如果碰到a-z之间的字�W�则停止�Q�如
          char s[]="HELLOkitty” ; // 注意: ,逗号在不 a-z之间
          sscanf( s, “%[^a-z]”, string ) ; // string=HELLO
      %*[^=] 前面�?* 可��C�Z��保存变量,跌��W�合条�g的字�W�串�?br>          char s[]="notepad=1.0.0.1001" ;
          char szfilename [32] = "" ;
          int i = sscanf( s, "%*[^=]", szfilename ) ; // szfilename=NULL,因�ؓ没保�?br>          int i = sscanf( s, "%*[^=]=%s", szfilename ) ; // szfilename=1.0.0.1001

patriking 2011-01-15 19:54 发表评论