1120. Friend Numbers (20)

本文介绍了一种算法,用于计算并统计一组整数中不同友数ID的数量及其具体值。友数是指两个整数如果其各位数字之和相同,则这两个数互为友数,该和即为它们的友数ID。通过输入一系列整数,程序能够输出不同友数ID的数量及这些ID从小到大的排序。

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1120. Friend Numbers (20)
时间限制 400 ms 内存限制 65536 kB 代码长度限制 16000 B
判题程序 Standard 作者 CHEN, Yue

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 frind 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
Sample Output:
4
3 6 9 26

#define _CRT_SECURE_NO_WARNINGS
#include <algorithm>
#include <iostream>
#include <cstring>
#include <iomanip>
#include <vector>
#include <string>
#include <cfloat>
#include <queue>
#include <map>
#include <set>

using namespace std;
const int MaxN = 10010;
bool isFriend[MaxN] = { 0 };
set<int> fridid;

int main()
{
#ifdef _DEBUG
    freopen("data.txt", "r+", stdin);
#endif // _DEBUG
    std::ios::sync_with_stdio(false);

    int N, num;
    cin >> N;
    for (int i = 0; i < N; ++i)
    {
        cin >> num;
        int frid = 0;
        while (num)
        {
            frid += num % 10;
            num /= 10;
        }

        if (!isFriend[frid])
        {
            isFriend[frid] = true;
            fridid.insert(frid);
        }
    }

    cout << fridid.size() << endl;
    for (auto it = fridid.begin(); it != fridid.end(); )
    {
        cout << *it;
        ++it;
        if (it != fridid.end())cout << " ";
    }


    return 0;
}
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