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PAT 1120 Friend Numbers[簡單]

怎麽 nts bottom code HERE pac all friend 現在

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 10?4??.

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

題目大意:兩個數如果每位相加結果相等,那麽結果就是個友好數,現在給出幾個數,求他們的友好數。

//就算一個數沒有和別的單位數和一樣的,也將這個算作友好數。

//我的AC

#include <iostream>
#include <map>
using namespace std;

int main() {
    int n;
    map<int,int> mp;
    cin>>n;
    int
t; for(int i=0;i<n;i++){ cin>>t; int x,sum=0; while(t!=0){ x=t%10; t/=10; sum+=x; } mp[sum]=1; sum=0; } cout<<mp.size()<<\n; for(auto it=mp.begin();it!=mp.end();){ cout<<it->first;//每次都不知道這裏的空格該怎麽控制?!!! if(++it!=mp.end()) cout<<" "; } return 0; }

//還是比較簡單的,就是用map;

在解決最後輸出格式的問題上,可以使用一個if判斷,註意應該是++it去判斷,而不是it++,深刻地理解了這個前+和後+的區別。

PAT 1120 Friend Numbers[簡單]