Map

Maps are a kind of associative containers that stores elements formed by the combination of a key value and a mapped value.

In a map, the key value is generally used to uniquely identify the element, while the mapped value is some sort of value associated to this key. Types of key and mapped value may differ. For example, a typical example of a map is a telephone guide where the name is the key and the telephone number is the mapped value.

Internally, the elements in the map are sorted from lower to higher key value following a specific strict weak ordering criterion set on construction.

As associative containers, they are especially designed to be efficient accessing its elements by their key (unlike sequence containers, which are more efficient accessing elements by their relative or absolute position).

Therefore, the main characteristics of a map as an associative container are:

• Unique key values: no two elements in the map have keys that compare equal to each other. For a similar associative container allowing for multiple elements with equivalent keys, see multimap.
• Each element is composed of a key and a mapped value. For a simpler associative container where the element value itself is its key, see set.
• Elements follow a strict weak ordering at all times. Unordered associative arrays, like unordered_map, are available in implementations following TR1.

Maps are also unique among associative containers in that the implement the direct access operator (operator[]) which allows for direct access of the mapped value.

In their implementation in the C++ Standard Template Library, map containers take four template parameters:

template < class Key, class T, class Compare = less<Key>, class Allocator = allocator<pair<const Key,T> > > class map;

Where the template parameters have the following meanings:

• Key: Type of the key values. Each element in a map is uniquely identified by its key value.
• T: Type of the mapped value. Each element in a map is used to store some data as its mapped value.
• Compare: Comparison class: A class that takes two arguments of the key type and returns a bool. The expression comp(a,b), where comp is an object of this comparison class and a and b are key values, shall return true if a is to be placed at an earlier position than b in a strict weak ordering operation. This can either be a class implementing a function call operator or a pointer to a function (see constructor for an example). This defaults to less<Key>, which returns the same as applying the less-than operator (a<b).
  The map object uses this expression to determine the position of the elements in the container. All elements in a map container are ordered following this rule at all times.
• Allocator: Type of the allocator object used to define the storage allocation model. By default, the allocator class template is used, which defines the simplest memory allocation model and is value-independent.

This container class supports bidirectional iterators.
Iterators to elements of map containers access to both the key and the mapped value. For this, the class defines what is called its value_type, which is a pair class with its first value corresponding to the const version of the key type (template parameter Key) and its second value corresponding to the mapped value (template parameter T):

typedef pair<const Key, T> value_type;

Iterators of a map container point to elements of this value_type. Thus, for an iterator called it that points to an element of a map, its key and mapped value can be accessed respectivelly with:

map<Key,T>::iterator it; (*it).first; // the key value (of type Key) (*it).second; // the mapped value (of type T) (*it); // the "element value" (of type pair<const Key,T>)

Naturally, any other direct access operator, such as -> or [] can be used, for example:

it->first; // same as (*it).first (the key value) it->second; // same as (*it).second (the mapped value)

Member functions

Iterators:

begin

Return iterator to beginning (public member function)

end	Return iterator to end (public member function)

rbegin

Return reverse iterator to reverse beginning (public member function)

rend	Return reverse iterator to reverse end (public member function)

Capacity:

empty

Test whether container is empty (public member function)

size	Return container size (public member function)

max_size

Return maximum size (public member function)

Element access:

operator[]

Access element (public member function)

Modifiers:

insert

Insert element (public member function)

erase

Erase elements (public member function)

swap	Swap content (public member function)

clear

Clear content (public member function)

Observers:

key_comp

Return key comparison object (public member function)

value_comp

Return value comparison object (public member function)

Operations:

find	Get iterator to element (public member function)

count

Count elements with a specific key (public member function)

lower_bound

Return iterator to lower bound (public member function)

upper_bound

Return iterator to upper bound (public member function)

equal_range

Get range of equal elements (public member function)

Allocator:

get_allocator

Get allocator (public member function)

Member types

member type	definition
key_type	Key
mapped_type	T
value_type	pair<const Key,T>
key_compare	Compare
value_compare	Nested class to compare elements (see member function value_comp)
allocator_type	Allocator
reference	Allocator::reference
const_reference	Allocator::const_reference
iterator	Bidirectional iterator
const_iterator	Constant bidirectional iterator
size_type	Unsigned integral type (usually same as size_t)
difference_type	Signed integral type (usually same as ptrdiff_t)
pointer	Allocator::pointer
const_pointer	Allocator::const_pointer
reverse_iterator	reverse_iterator<iterator>
const_reverse_iterator	reverse_iterator<const_iterator>

3 3
1 2
1 3
2 3
3 2
1 2
2 3
0

1
0

网上说要用并查集,唉~~不懂啊,算法白痴,按自己的思路做,欧拉回路每个节点必有偶数个度(通路可以有0个或2个为基数),基本思路就这样

 1#include<iostream>
 2using namespace std;
 3int d[1001],set[1001],a,b;
 4int main()
 5{
 6    int n,m,flag;
 7    while(cin>>n&&n)
 8    {
 9        for(int i=1;i<=n;i++)
10        {
11            d[i]=0;
12            set[i]=i;
13        }
14        cin>>m;
15        flag=0;        
16        while(m--)
17        {
18            cin>>a>>b;
19            if(set[a]>set[b])    //判断是否连通用
20                set[a]=set[b];
21            else
22                set[b]=set[a];
23            d[a]++;
24            d[b]++;
25        }
26        for(i=1;i<=n;i++)    //每个节点都必有偶数个度
27            if(d[i]&1)
28            {
29                flag=1;
30                break;
31            }
32        if(!flag)
33            for(i=1;i<=n;i++)
34                if(set[i]!=1)
35                {
36                    flag=1;
37                    break;
38                }
39        if(flag)
40            cout<<"0"<<endl;
41        else
42            cout<<"1"<<endl;
43    }
44    return 0;
45}

第一行一个整数 T 代表case数。
每个case第一行是两个整数 n（1≤ n≤ 100000）, p( 1≤p≤1000000),第二行是n个整数
1≤a(k) ≤10000( 1≤k≤n).

Output

每个case输出一行，存在连续个a，这些a的和能被p整除就输出“Yes”,否则输出“No”.

Sample Input

2
3 5
1 3 2
3 5
4 4 4

Sample Output

Yes
No

#include<stdio.h>
long int b[100000]={0};
int main()
{
    long int t,n,p,a;
    long int c,i,j;
    scanf("%d",&t);
    while(t--)
    {
        c=0;
        scanf("%d %d",&n,&p);
        for(i=0;i<n;i++)
        {
            scanf("%d",&a);
            a%=p;
            if(!c)
            {
                for(j=i;j>=0;j--)
                {
                    if(!c)
                    {
                        b[j]=b[j-1]+a;
                        if(b[j]%p==0)
                            c=1;
                    }
                }
                b[0]=a;
                if(b[0]%p==0)
                    c=1;
            }
        }
        if(c)
            printf("Yes\n");
        else
            printf("No\n");
    }
    return 0;
}

#include<iostream>
long int b[1000000];
int main()
{
    long int t,n,p,a,s;
    long int c,i;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%ld %ld",&n,&p);
        c=0;
        s=0;        
        memset(b,0,p*sizeof(long int));
        for(i=0;i<n;i++)
        {
            scanf("%ld",&a);
            s+=a%p;
            if(!c)
            {
                if(b[s%p]==s%p)
                    c=1;
                else
                    b[s%p]=s%p;
            }
        }
        if(c)
            printf("Yes\n");
        else
            printf("No\n");
    }
    return 0;
}