完全数计算

题目描述：https://www.nowcoder.com/practice/7299c12e6abb437c87ad3e712383ff84?tpId=37&tags=&title=&diffculty=0&judgeStatus=0&rp=1&ru=/ta/huawei&qru=/ta/huawei/question-ranking

参考：https://www.nowcoder.com/profile/1470346/codeBookDetail?submissionId=12557587

#include<iostream>

using namespace std;

bool is_Perfect_Num(int n)
{
    int sum=0,j=1,t=n/2;
    while(j<t)
    {
        if(n%j==0)
        {
            sum+=j;
            if(j!=1)
            {
                t = n/j;
                if(j!=n/j) sum+=n/j;
            }
        }
        ++j;
        if(sum>n)  return false;
    }
    return (sum==n)?true:false;
}
int count_Perfect_Num(int n)
{
    if(n<=0||n>=500000) return -1;
    int cnt = 0;
    for(int i=1;i<=n;i++)
    {
        if(is_Perfect_Num(i))++cnt;
    }
    return cnt;
}
int main()
{
    int num;
    while(cin>>num)
    {
        cout<<count_Perfect_Num(num)<<endl;
    }

    return 0;
}

解法2：https://www.nowcoder.com/profile/7562989/codeBookDetail?submissionId=12524988

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

int count(int n) {
    int cnt = 0;
    if ((n < 0) | (n>500000))
        return -1;
    else {
        for (int i = 2; i <= n; i++) {
            int sum = 0;
            int sq = sqrt(i);
            for (int j = 2; j <= sq; j++) {
                if (i%j == 0) {
                    if (i / j == j)
                        sum += j;
                    else
                        sum += j + (i / j);
                }
            }
            if (sum + 1 == i)
                cnt++;
        }
        return cnt;
    }
}

int main() {
    int input;
    while(cin>>input)
        cout << count(input) << endl;
    return 0;
}