c184. 盈虧互補

給一個數字 n
判斷他是不是完全數
輸出計算過程
如果是完全數，輸出 {n} is perfect.
如果不是 且 有友好數，輸出 {n} and {m} are friends.
如果不是 且 沒有友好數，輸出 {n} has no friends.

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解法一

✅ 完整代碼

評分結果(參考) ： AC (2ms, 348KB)

#include<bits/stdc++.h>
#define int long long
using namespace std;

int getFriend(int num){
    if(num == 1){
        cout << "=0\n";
        return 0;
    }
    
    int sum = 1;
    set<int> st;
    for(int i=2; i<=sqrt(num); i++){
        if(num%i == 0){
            st.insert(i);
            st.insert(num/i);
        }
    }
    
    cout << "1";
    for(int i : st){
        cout << "+" << i;
        sum += i;
    }
    cout << "=" << sum << "\n";
    
    return sum;
}

signed main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    
    int n;
    cin >> n;
    
    int x = getFriend(n);
    if(x == n){
        cout << n << " is perfect.";
    }
    else{
        int y = getFriend(x);
        if(y == n) cout << n << " and " << x << " are friends.";
        else cout << n << " has no friends.";
    }
    
    return 0;
}