java关于friendly

Posted 2023-04-23

这是题目：
用public、private、protected以及“友好的”数据成员及方法成员创建一个类。创建属于这个类的一个对象，并观察在试图访问所有类成员时会获得哪种类型的编译器错误提示。注意同一个目录内的类属于“默认”包的一部分。

下面是我写的代码：
class Test
public int a;
private int b;
protected int c;
int d;

public void a()
System.out.println("public" + a);


private void b()
System.out.println("private" + b);


protected void c()
System.out.println("protected" + c);


void d()
System.out.println("friendly" + d);




public class Test01
public static void main(String[] args)
Test test = new Test();
test.a();
//test.b();//私有变量在Test01类中不可见
test.c();
test.d();
System.out.println(test.a);
//System.out.println(test.b); //私有方法在Test01类中不可见
System.out.println(test.c);
System.out.println(test.d);



我想请问一下，为什么我把
class Test
public int a;
private int b;
protected int c;
int d;
这里面的d定义成friendly int d;
编译就不能通过。。
这是为什么呀。
friendly是默认的意思呀，为什么加了就不能通过了

java中只有public、private、protected、default这几种修饰符，没有friendly修饰符，没加修饰符就是friendly。friendly只是一种说法，把它认为是default，即默认的就好！！ 参考技术A friendly两种情况，友元函数和友元类， 没有友元成员变量这回事

HDU 5305 Friends (DFS)

Friends

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

 Total Submission(s): 163    Accepted Submission(s): 61

Problem Description

There are n people and m pairs of friends. For every pair of friends, they can choose to become online friends (communicating using online applications) or offline friends (mostly using face-to-face communication). However, everyone in these n people wants to have the same number of online and offline friends (i.e. If one person has x onine friends, he or she must have x offline friends too, but different people can have different number of online or offline friends). Please determine how many ways there are to satisfy their requirements.

 

Input

The first line of the input is a single integer T (T=100), indicating the number of testcases.

For each testcase, the first line contains two integers n (1≤n≤8) and m (0≤m≤n(n?1)2), indicating the number of people and the number of pairs of friends, respectively. Each of the next m lines contains two numbers x and y, which mean x and y are friends. It is guaranteed that x≠y and every friend relationship will appear at most once.

 

Output

For each testcase, print one number indicating the answer.

 

Sample Input

2
3 3
1 2
2 3
3 1
4 4
1 2
2 3
3 4
4 1

 

Sample Output

0
2

 

Source

2015 Multi-University Training Contest 2

题目大意：一些朋友关系可离线可在线，要求每一个人的离线朋友数等于在线朋友数

题目分析：数据量不大。直接爆搜。记录每一个点的度数，奇数的直接不可能，偶数的分成两个数组，c1[i]表示i的在线朋友数，c2[i]表示i的离线朋友数，然后一条边一条边搜即可了，注意一个剪枝。当一条边的两个端点有一个c值为0，则return

#include <cstdio>
#include <cstring>
bool mp[10][10];
int deg[10], c1[10], c2[10];
int n, m, cnt, ans;

struct EDGE
{
    int u, v;
}e[100000];

void DFS(int cur)
{
    if(cur == m + 1)
    {
        ans ++;
        return;
    }
    int v = e[cur].v;
    int u = e[cur].u;
    if(c1[u] && c1[v])
    {
        c1[u] --;
        c1[v] --;
        DFS(cur + 1);
        c1[u] ++;
        c1[v] ++;
    }
    if(c2[u] && c2[v])
    {
        c2[u] --;
        c2[v] --;
        DFS(cur + 1);
        c2[u] ++;
        c2[v] ++;
    }
    return;
}

int main()
{
    int T;
    scanf("%d", &T);
    while(T --)
    {
        cnt = 0;
        ans = 0;
        memset(deg, 0, sizeof(deg));
        memset(e, 0, sizeof(e));
        memset(c1, 0, sizeof(c1));
        memset(c2, 0, sizeof(c2));
        scanf("%d %d", &n, &m);
        for(int i = 1; i <= m; i++)
        {
            scanf("%d %d", &e[i].u, &e[i].v);
            deg[e[i].u] ++;
            deg[e[i].v] ++;
        }
        bool f = false;
        for(int i = 1; i <= n; i++)
        {
            c1[i] = c2[i] = deg[i] / 2;
            if(deg[i] & 1)
            {
                f = true;
                break;
            }
        }
        if(f)
        {
            printf("0\n");
            continue;
        }
        DFS(1);
        printf("%d\n", ans);
    }
}