PAT (Advanced Level) Practice 1120 Friend Numbers（20 分）

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最新推荐文章于 2020-02-15 11:13:32 发布

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博客围绕Friend Numbers问题展开，定义两个整数若各位数字之和相同则为“朋友数”，该和为“朋友ID”。给出输入输出规范及示例，解题思路是处理每个数，用set记录朋友ID并遍历输出，以计算给定整数中不同朋友ID的数量。

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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

题意：定义一个数的每位数之和为 friend id ；问数组中有哪几个friend id

思路：处理每个数，set记录friend id，遍历输出

代码：

#include <bits/stdc++.h>
using namespace std;
int check(int x)
{
    int sum=0;
    while(x)
    {
        sum+=x%10;
        x/=10;
    }
    return sum;
}
set<int>s;
int main()
{
    int n;scanf("%d",&n);
    int x;
    for(int i=0;i<n;i++)
    {
        scanf("%d",&x);
        s.insert(check(x));
    }
    printf("%d\n",s.size());
    set<int>::iterator it;
    for(it=s.begin();it!=s.end();it++)
    {
        if(it!=s.begin())printf(" ");
        cout<<*it;
    }
    printf("\n");
}