1120 Friend Numbers（20 分）

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博客围绕Friend Numbers问题展开，介绍了问题描述、输入输出规范及示例。分析指出接收输入数据时将数字各位相加，结果插入set集合，set的size值即不同friend ID个数。还分享了收获，如auto的便捷使用、求整数每位数字和的两种方法及set默认排序规则。

不是自己做的，优快云现在不让转载，只能复制了，标明出处：https://www.liuchuo.net/archives/2901

1120 Friend Numbers（20 分）

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 frind ID’s among them.
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

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

收获：1.auto用起来的真的很方便
2.整数每位数字和，两种方法：（1）代码中的数学方法
（2）转换成char相加后再把和转为int
3.set< int>中默认按从小到达排列

#include <cstdio>
#include <set>
using namespace std;
int getFriendNum(int num) {
    int sum = 0;
    while(num != 0) {
        sum += num % 10;
        num /= 10;
    }
    return sum;
}
int main() {
    set<int> s;
    int n, num;
    scanf("%d", &n);
    for(int i = 0; i < n; i++) {
        scanf("%d", &num);
        s.insert(getFriendNum(num));
    }
    printf("%d\n", s.size());
    for(auto it = s.begin(); it != s.end(); it++) {
        if(it != s.begin()) printf(" ");
        printf("%d", *it);
    }
    return 0;
}