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PAT 1120. Friend Numbers (20) Java

Two integers are called “friend numbers” if they share the same sum of their digits, and the sum is their “friend ID”. For example, 123 and 51 are friend numbers since 1+2+3 = 5+1 = 6, and 6 is their friend ID. Given some numbers, you are supposed to count the number of different friend ID’s among them. Note: a number is considered a friend of itself.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N. Then N positive integers are given in the next line, separated by spaces. All the numbers are less than 10⁴.

Output Specification:

For each case, print in the first line the number of different friend ID’s among the given integers. Then in the second line, output the friend ID’s in increasing order. The numbers must be separated by exactly one space and there must be no extra space at the end of the line.

Sample Input:

8
123 899 51 998 27 33 36 12

1 2	8 123 899 51 998 27 33 36 12

Sample Output:

4
3 6 9 26

1 2	4 3 6 9 26

分析：在接收输入数据的时候就把该数字的每一位相加，并把结果插入一个set集合中。因为set是有序的、不重复的，所以set的size值就是输出的个数，set中的每一个数字即所有答案的数字序列

此为Java版，C/C++版本请戳：http://www.liuchuo.net/archives/2901

import java.util.Scanner;
import java.util.Set;
import java.util.TreeSet;

public class Main {
	public static void main(String[] args) {
		Scanner scanner = new Scanner(System.in);
		int numOfInteger = scanner.nextInt();
		Set<Integer> set = new TreeSet<Integer>();
		for (int i = 0; i < numOfInteger; i++) {
			int temp = scanner.nextInt();
			set.add(getFriendId(temp));
		}
		scanner.close();
		System.out.println(set.size());
		boolean isFirst = true;
		for (int id : set) {
			if (!isFirst) {
				System.out.print(" ");
			}
			System.out.print(id);
			isFirst = false;
		}
	}
	
	private static int getFriendId(int num) {
		int sum = 0;
		while(num != 0) {
			sum += num % 10;
			num /= 10;
		}
		return sum;
	}
}

import java.util.Scanner;

import java.util.Set;

import java.util.TreeSet;

public class Main {

public static void main(String[] args) {

Scanner scanner = new Scanner(System.in);

int numOfInteger = scanner.nextInt();

Set<Integer> set = new TreeSet<Integer>();

for (int i = 0; i < numOfInteger; i++) {

int temp = scanner.nextInt();

set.add(getFriendId(temp));

}

scanner.close();

System.out.println(set.size());

boolean isFirst = true;

for (int id : set) {

if (!isFirst) {

System.out.print(" ");

}

System.out.print(id);

isFirst = false;

}

private static int getFriendId(int num) {

int sum = 0;

while(num != 0) {

sum += num % 10;

num /= 10;

}

return sum;

}

1121. Damn Single (25)-PAT甲级真题

“Damn Single (单身狗)” is the Chinese nickname for someone who is being single. You are supposed to find those who are alone in a big party, so they can be taken care of.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (<=50000), the total number of couples. Then N lines of the couples follow, each gives a couple of ID’s which are 5-digit numbers (i.e. from 00000 to 99999). After the list of couples, there is a positive integer M (<=10000) followed by M ID’s of the party guests. The numbers are separated by spaces. It is guaranteed that nobody is having bigamous marriage (重婚) or dangling with more than one companion.

Output Specification:

First print in a line the total number of lonely guests. Then in the next line, print their ID’s in increasing order. The numbers must be separated by exactly 1 space, and there must be no extra space at the end of the line.

Sample Input:

3
11111 22222
33333 44444
55555 66666
7
55555 44444 10000 88888 22222 11111 23333

Sample Output:

5
10000 23333 44444 55555 88888

分析：设立数组couple[i] = j表示i的对象是j。一开始先设置为都是-1。设立数组isExist表示某人的对象是否来到了派对上。接收数据的时候，对于每一对a和b，将couple的a设置为b，b设置为a，表示他俩是一对。对于每一个需要判断的人，将其存储在guest数组里面，如果它不是单身的（也就是如果它的couple[guest[i]] != -1）那么就将它对象的isExist设置为1，表示他对象的对象（也就是他自己）来到了派对。这样所有isExist不为1的人，对象是没有来到派对的。把所有的人遍历后插入一个集合set里面，set的size就是所求的人数，set里面的所有数就是所求的人的递增排列～～～

#include <iostream>
#include <set>
#include <vector>
using namespace std;
int main() {
    int n, a, b, m;
    scanf("%d", &n);
    vector<int> couple(100000);
    for (int i = 0; i < 100000; i++)
        couple[i] = -1;
    for (int i = 0; i < n; i++) {
        scanf("%d%d", &a, &b);
        couple[a] = b;
        couple[b] = a;
    }
    scanf("%d", &m);
    vector<int> guest(m), isExist(100000);
    for (int i = 0; i < m; i++) {
        scanf("%d", &guest[i]);
        if (couple[guest[i]] != -1)
            isExist[couple[guest[i]]] = 1;
    }
    set<int> s;
    for (int i = 0; i < m; i++)
        if (!isExist[guest[i]]) s.insert(guest[i]);
    printf("%d\n", s.size());
    for (auto it = s.begin(); it != s.end(); it++) {
        if (it != s.begin()) printf(" ");
        printf("%05d", *it);
    }
    return 0;
}

#include <iostream>

#include <set>

#include <vector>

using namespace std;

int main() {

int n, a, b, m;

scanf("%d", &n);

vector<int> couple(100000);

for (int i = 0; i < 100000; i++)

couple[i] = -1;

for (int i = 0; i < n; i++) {

scanf("%d%d", &a, &b);

couple[a] = b;

couple[b] = a;

}

scanf("%d", &m);

vector<int> guest(m), isExist(100000);

for (int i = 0; i < m; i++) {

scanf("%d", &guest[i]);

if (couple[guest[i]] != -1)

isExist[couple[guest[i]]] = 1;

}

set<int> s;

for (int i = 0; i < m; i++)

if (!isExist[guest[i]]) s.insert(guest[i]);

printf("%d\n", s.size());

for (auto it = s.begin(); it != s.end(); it++) {

if (it != s.begin()) printf(" ");

printf("%05d", *it);

}

return 0;

}

【iOS】Illegal Configuration: The Label outlet from the ViewController to the UILabel is invalid. Outlets cannot be connected to repeating content.

报错：Illegal Configuration: The Label outlet from the ViewController to the UILabel is invalid. Outlets cannot be connected to repeating content.

产生原因：tableview里面的label是动态的，而用storyboard那样拖动绑定过来的outlet是静态的，所以不能采用这种方法来绑定。应该采用如下方法：

首先给label设置一个Tag，比如105

snip20161208_198

然后在代码中使用viewWithTag的方法访问：

public func tableView(_ tableView: UITableView, cellForRowAt indexPath: IndexPath) -> UITableViewCell {
        let cell = self.tableView.dequeueReusableCell(withIdentifier: "cell")! as UITableViewCell
        let myLabel = cell.viewWithTag(105) as! UILabel
        myLabel.text = myArr[indexPath.row]  
        return cell
    }