Problem F: STL——集合运算

Problem F: STL——集合运算

Time Limit: 1 Sec  Memory Limit: 128 MB

Submit: 3767  Solved: 1927

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Description

集合的运算就是用给定的集合去指定新的集合。设A和B是集合，则它们的并差交补集分别定义如下：

A∪B={x|x∈A∨x∈B}

A∩B={x|x∈A∧x∈B}

A-B={x|x∈A∧x不属于 B}

SA ={x|x∈(A∪B)∧x 不属于A}

SB ={x|x∈(A∪B)∧x 不属于B}

Input

第一行输入一个正整数T，表示总共有T组测试数据。（T<=200）

然后下面有2T行，每一行都有n+1个数字，其中第一个数字是n(0<=n<=100)，表示该行后面还有n个数字输入。

Output

对于每组测试数据，首先输出测试数据序号，”Case #.NO”，

接下来输出共7行，每行都是一个集合，

前2行分别输出集合A、B，接下5行来分别输出集合A、B的并(A u B)、交(A n B)、差(A – B)、补。

集合中的元素用“{}”扩起来，且元素之间用“， ”隔开。

Sample Input

14 1 2 3 10

Sample Output

Case# 1:A = {1, 2, 3}B = {}A u B = {1, 2, 3}A n B = {}A - B = {1, 2, 3}SA = {}SB = {1, 2, 3}

HINT

如果你会用百度搜一下关键字“stl set”，这个题目我相信你会很快很轻松的做出来。加油哦！

Append Code

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#include <iostream>
#include <algorithm>
#include <set>
using namespace std;
int main()
{
    int T, j, n, a[110];
    cin >> T;
    for(int i = 1; i <= T; i++)
    {
        set <int> A; set <int> B; set <int> C;
        A.clear(); B.clear(); C.clear(); set <int>:: iterator set_iter;
        cin >> n;
        for( j = 0; j < n; ++j)
           cin >> a[j];
        A.insert(a, a+n);
        cin >> n;
        for(int j = 0; j < n; j++)
           cin >> a[j];
        B.insert(a, a + n);
        cout << "Case# " << i << ":" << endl;
        cout << "A = {";
        for(set_iter = A.begin(); set_iter != A.end() ; ++set_iter) {
                if(set_iter != A.begin())
                  cout << ", ";
            cout << *set_iter;
        }
        cout << "}" <<endl;
        cout << "B = {";
        for(set_iter = B.begin(); set_iter != B.end() ; ++set_iter) {
                if(set_iter != B.begin())
                  cout << ", ";
            cout << *set_iter;
        }
        cout << "}" <<endl;
        cout << "A u B = {";
        set_union(A.begin(), A.end(), B.begin(), B.end(), insert_iterator< set<int > >(C,C.begin()));
        for(set_iter = C.begin(); set_iter != C.end() ; ++set_iter) {
                if(set_iter != C.begin())
                  cout << ", ";
            cout << *set_iter;
        }
        cout << "}" <<endl;
        C.clear();
        set_intersection( A.begin(), A.end(), B.begin(), B.end(), insert_iterator< set<int> >(C, C.begin()));
        cout << "A n B = {";
         for(set_iter = C.begin(); set_iter != C.end() ; ++set_iter) {
                if(set_iter != C.begin())
                  cout << ", ";
            cout << *set_iter;
        }
         cout << "}" <<endl;
         C.clear();
         cout << "A - B = {";
         set_difference(A.begin(), A.end(), B.begin(), B.end(), insert_iterator< set<int> >(C, C.begin()));
         for(set_iter = C.begin(); set_iter != C.end(); set_iter++ )
         {
             if(set_iter != C.begin())
                cout << ", ";
             cout << *set_iter;
         }
         cout << "}" << endl;
          C.clear(); set<int> C2;
         cout << "SA = {";
         set_union(A.begin(),A.end(),B.begin(),B.end(),insert_iterator<set<int> >(C2,C2.begin()));
         set_difference(C2.begin(), C2.end(), A.begin(), A.end(), insert_iterator< set<int> >(C, C.begin()));
         for(set_iter = C.begin(); set_iter != C.end(); set_iter++ )
         {
             if(set_iter != C.begin())
                cout << ", ";
             cout << *set_iter;
         }
         cout << "}" << endl;
          C.clear();
         cout << "SB = {";
         set_difference(C2.begin(), C2.end(), B.begin(), B.end(), insert_iterator< set<int> >(C, C.begin()));
         for(set_iter = C.begin(); set_iter != C.end(); set_iter++ )
         {
             if(set_iter != C.begin())
                cout << ", ";
             cout << *set_iter;
         }
         cout << "}" << endl;
         A.clear(); B.clear(); C.clear(); C2.clear();
    }
}