"CCF" to find the opposite number

Topic 1 opposite numbers
Time limit: 1 seconds space limit: 256MB
Problem description
There are n non-0 integers that are not identical. Please make up a program to find out how many of them are opposite numbers (A and? A are pairs of opposite numbers).
Input format
The first line consists of a positive integer n. (1≤n≤500).
The second behavior n a non-0 integer separated by a single space, the absolute value of each number does not exceed 1000, guaranteeing that the integers are different.
Output format
Only one integer is output, that is, how many pairs of opposite numbers are included in the N number.
Input sample
5
123-1-2
Output sample

2

① Linked List method

#include <iostream> #include <time.h> #include <stdlib.h>using namespace Std;class node{public:int Data Node* Next; Node () {next = NULL;}}; Class linkedlist{private:node* head; node* last;int len;p ublic:linkedlist () {head = Null;last = Null;len = 0;} ~linkedlist () {node*p=head; Node*q;while (p) {q = P;p=p->next;delete q;}} void Add (int element) {if (head) {last->next=new Node (); last->next->data=element;last = Last->next;len + +;} Else{head = new Node (); head->data = Element;last = Head;len + +;}} void Remove (int i) {node* p=head; node* q=null;if (i==0) {Q=p;head=head->next;delete Q;return;} while (i--) {q=p;p=p->next;} if (p->next) {node* m = P;p=p->next;q->next = P;delete m;} Else{last = Q;last->next=null;delete p;}} void display () {node* p = head;while (p) {cout<<p->data<< "";p =p->next;} Cout<<endl;} node* GetHead () {return head;}; int main () {LinkedList pn;//positive linkedlist nn;//negative srand ((unsigned int) time (0)); int Tn,n;cin>>tn;while (tn--) {n= Rand()%1000 * ((rand ()%2==0) 1:-1);cout<<n<< "";//cin>>n;if (n>0) {pn.add (n);} Else{nn.add (-n);}} cout<<endl; node* pn_head = Pn.gethead (); node* nn_head;int I,amout=0;while (pn_head) {nn_head = Nn.gethead (); i = 0;while (nn_head) {if (nn_head->data==pn_head- >data) {cout<<nn_head->data<< "-" <<pn_head->data<<endl;amout++;nn.remove (i); break;} I++;nn_head=nn_head->next;} Pn_head=pn_head->next;} Cout<<amout<<endl;return 0;}

② Collection Method (reference address: http://zhidao.baidu.com/link?url= govljwhisgnjhcyu39b7uupe7yb95qxtrviqtpctpzzls-umzfjshm0tw2sxhtrd895mj8p7mwopphtqzujpt7pvtevwpyufv_oueezkf7k)

#include <iostream> #include <set>using namespace Std;int main () {int n,n;set<int> s;cin>>n;int i = N;while (i--) {cin>>n;if (n>=0) {S.insert (N);} Else{s.insert (-n);}} cout<< (N-s.size ()) <<endl;return 0;}

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"CCF" to find the opposite number