3 条题解

  • 0
    @ 2024-08-28 12:18:13
    #include<iostream>
    using namespace std;
    int a[10010], b[10010];
    int main(){
        ios::sync_with_stdio(false);
        int a1, b1;
        cin >> a1;
        for(int i = 1; i <= a1; i++)cin >> a[i];
        cin >> b1;
        for(int i = 1; i <= b1; i++)cin >> b[i];
        for(int i = 1; i <= a1; i++){
            for(int j = 1; j <= b1; j++){
                if(a[i] == b[j])cout << a[i] << ' ';
            }
        }
        return 0;
    }
    
  • 0
    @ 2021-04-01 21:13:37
    #include<iostream>
    #include<vector>
    using namespace std;
    int main()
    {
        vector <int> a,b;
        int na,nb,j,j1;cin>>na;
        for(int i=0;i<na;i++)
        {
            cin>>j;
            a.push_back(j);
        }
        cin>>nb;
        for(int i=0;i<nb;i++)
        {
            cin>>j1;
            b.push_back(j1);
        }
        for(int i=0;i<a.size();i++)
            for(int j=0;j<b.size();j++)
                if(a[i]==b[j])
                    cout<<a[i]<<" ";
        return 0;
    } 
    
  • -1
    @ 2019-03-31 13:28:34
    #include <iostream>
    using namespace std;
    class SeqList
    {
        int a[205], n;
    public:
        SeqList()
        {
    
        }
        void init(int b[], int n)
        {
            this->n = n;
            for (int i = 0; i < n; i++)
            {
                a[i] = b[i];
            }
        }
        int Search(int x)
        {
            for (int i = 0; i < n; i++)
            {
                if (a[i] == x)
                    return 1;
            }
            return -1;
        }
        int *Inter(SeqList Lb)
        {
            int *b = new int[n + Lb.n];
            int index = 0;
            for (int i = 0; i < n; i++)
            {
                if (Lb.Search(a[i]) == 1)
                {
                    b[index] = a[i];
                    index++;
                }
            }
            for (int i = 0; i < index; i++)
            {
                cout << b[i] << " ";
            }
            return b;
        }
        void print()
        {
            for (int i = 0; i < n; i++)
                cout << a[i] << " ";
        }
    };
    int main()
    {
        int na; cin >> na;
        int b[205];
        for (int i = 0; i < na; i++)
        {
            cin >> b[i];
        }
        SeqList La;
        La.init(b, na);
    
        int nb; cin >> nb;
        int c[205];
        for (int i = 0; i < nb; i++)
        {
            cin >> c[i];
        }
        SeqList Lb;
        Lb.init(c, nb);
        int *d = new int[na + nb];
        d = La.Inter(Lb);
        return 0;
    }
    
  • 1

OO3-2 无序顺序表类的集合运算

信息

ID
1015
难度
3
分类
(无)
标签
递交数
325
已通过
150
通过率
46%
被复制
7
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