Problem B: STL——集合运算

Time Limit: 1 Sec Memory Limit: 128 MB
Submit: 2547 Solved: 1329
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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！

#include <iostream>
#include <algorithm>
#include <set>

using namespace std;
void print(const set<int> &t)
{
    set<int>::iterator it;
    cout << "{";
    for(it = t.begin(); it != t.end(); it++)
        if(it == t.begin()) cout << *it;
        else cout << ", " << *it;
    cout << "}" << endl;
}
int main()
{

    int t;
    cin>>t;
    int i=0;
    while(t--)
    {
        cout<<"Case# "<<++i<<":"<<endl;
        set<int>s1;
        set<int>s2;
        set<int>tmp1,tmp2,tmp3,tmp4,tmp5;
        int m;
        cin>>m;
        for(int i=0;i<m;i++)
        {
            int tmp;
            cin>>tmp;
            s1.insert(tmp);
        }
        int k;
        cin>>k;
        for(int i=0;i<k;i++)
        {
            int tmp;
            cin>>tmp;
            s2.insert(tmp);
        }
        cout<<"A = ";
        print(s1);
        cout<<"B = ";
        print(s2);
        set_union(s1.begin(),s1.end(),s2.begin(),s2.end(),inserter(tmp1,tmp1.begin()));
        cout<<"A u B = ";
        print(tmp1);
        set_intersection(s1.begin(), s1.end(), s2.begin(), s2.end(), inserter(tmp2, tmp2.begin()));
        cout<<"A n B = ";
        print(tmp2);
        set_difference(s1.begin(), s1.end(), s2.begin(), s2.end(), inserter(tmp3, tmp3.begin()));
        cout<<"A - B = ";
        print(tmp3);
        set_difference(tmp1.begin(), tmp1.end(), s1.begin(), s1.end(), inserter(tmp4, tmp4.begin()));
        cout<<"SA = ";
        print(tmp4);
        set_difference(tmp1.begin(), tmp1.end(), s2.begin(), s2.end(), inserter(tmp5, tmp5.begin()));///set_??(a.begin(),a.end(),b.begin(),b.end(),inserter(x,x.begin());
        cout<<"SB = ";
        print(tmp5);

    }
    return 0;
}
/*
#define _CRT_SECURE_NO_WARNINGS

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <vector>
#include <cstring>

using namespace std;
const int maxn = 40000 + 10;
int a[maxn];
int b[maxn];

int main()
{
	int t;
	scanf("%d", &t);
	for (int kase = 1; kase <= t; kase++)
	{
		memset(a, 0, sizeof(a));
		memset(b, 0, sizeof(b));
		printf("Case #%d:\n", kase);
		int num1, num2;
		scanf("%d", &num1);
		for (int i = 0; i < num1; i++)
			scanf("%d", &a[i]);
		scanf("%d", &num2);
		for (int i = 0; i < num2; i++)
			scanf("%d", &b[i]);
		vector<int> aANDb, ab, a_b, b_a;
		for (int i = 0, j = 0; i < num1 || j < num2;)
		{
			if (i < num1 && j < num2 && a[i] == b[j])
			{
				aANDb.push_back(a[i]);
				ab.push_back(a[i]);
				i++, j++;
			}
			else if (i < num1 && j < num2 && a[i] < b[j] || j >= num2)
			{
				aANDb.push_back(a[i]);
				a_b.push_back(a[i]);
				i++;
			}
			else if (i < num1 && j < num2 && b[j] < a[i] || i >= num1)
			{
				aANDb.push_back(b[j]);
				b_a.push_back(b[j]);
				j++;
			}
		}
		int len = ab.size();
		for (int i = 0; i < len; i++)
		{
			if (i)
				printf(" ");
			printf("%d", ab[i]);
		}
		puts("");
		len = aANDb.size();
		for (int i = 0; i < len; i++)
		{
			if (i)
				printf(" ");
			printf("%d", aANDb[i]);
		}
		puts("");
		len = a_b.size();
		for (int i = 0; i < len; i++)
		{
			if (i)
				printf(" ");
			printf("%d", a_b[i]);
		}
		puts("");
		len = b_a.size();
		for (int i = 0; i < len; i++)
		{
			if (i)
				printf(" ");
			printf("%d", b_a[i]);
		}
		puts("");

	}
	return 0;
}*/

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Problem B: STL——集合运算