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  • poj1528

    简单题

    View Code
    #include<iostream>
    #include<iomanip>
    #include<cmath>
    #include<string>
    using namespace std;
    char level[3][20]={"DEFICIENT","PERFECT","ABUNDANT"};
    char a[1000];
    void output(int n)
    {
        int i;
        int mid=n/2;
        int sum=1;
        for(i=2;i<=mid;i++)
        {
            if(n%i==0)
            {
                sum+=i;
            }
        }
        if(sum<n)
            cout<<setw(5)<<n<<"  "<<level[0]<<endl;
        else if(sum==n)
            cout<<setw(5)<<n<<"  "<<level[1]<<endl;
        else cout<<setw(5)<<n<<"  "<<level[2]<<endl;
    }
    
    
    int main()
    {
        int i,j;
        int N;
        int cur;
        cout<<"PERFECTION OUTPUT"<<endl;
        while(cin>>cur&&cur!=0)
        {
            
            if(cur==0)
                break;
            if(cur==1)
            {
                cout<<setw(5)<<cur<<"  "<<level[0]<<endl;
            }
            else
                output(cur);
        }
        cout<<"END OF OUTPUT"<<endl;
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/rainydays/p/2866328.html
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