Programming Forums > Application Development > C++

adenus · Mar 19th, 2008, 3:29 PM

#include <iostream>
using namespace std;

//aprekjina funkcija

int Perfectnumber (int rob) {
    int result;
    int sum = 0;
        for (int i = 1; i < rob; i++) {
            if (rob % i == 0) {
               sum = sum + i;
            }
        }
        if (sum == rob) {
          result=sum;
        }
return (result);
}
//galvena funkcija
int main() {
int rez;
int lim;
    cout<<"ievadi robesu lidz kurai meklet : ";
          cin>>lim ;       
    for (int i = 1; i < lim; i++) {
        rez=Perfectnumber(lim);
    }
        cout<<"perfect numbers from 1 to"<<lim<<" : "<<rez<<endl;
    system ("pause");
return 0;
}

why dont it cout perfect numbers, ? and in one forum for me advice to inicialise result, but i dont know which value add to it?

adenus · Mar 19th, 2008, 3:59 PM

pls help!

peaceofpi · Mar 19th, 2008, 4:06 PM

Quote:

(Toggle Plain Text)

    for (int i = 1; i < lim; i++) {
        rez=Perfectnumber(lim);
    }
        cout<<"perfect numbers from 1 to"<<lim<<" : "<<rez<<endl;

Rethink the logic here.

grumpy · Mar 19th, 2008, 4:11 PM

Don't bump threads. If people have an answer to offer, they will. If you don't get a reply, it means people can't help or are disinclined to.

In any event, the code;

(Toggle Plain Text)

for (int i = 1; i < lim; i++) {
        rez=Perfectnumber(lim);
    }
cout<<"perfect numbers from 1 to"<<lim<<" : "<<rez<<endl;

only prints out the value of lim, and the value of rez produced in the last iteration through the for loop. If you want to print out every value that rez takes, put the output statement inside the loop.

adenus · Mar 19th, 2008, 4:24 PM

(Toggle Plain Text)

#include <iostream>
using namespace std;
//aprekjina funkcija

int Perfectnumber (int rob) {
    int result;
    int sum = 0;
        for (int i = 1; i < rob; i++) {
            if (rob % i == 0) {
               sum += i;
            }
            if (sum == rob) { 
               return true;
            } else {
               return false;       
            }
        }
return true;
}
//galvena funkcija
int main() {
int rez;
int lim;
    cout<<"ievadi robesu lidz kurai meklet : ";
          cin>>lim ;       
    for (int i = 1; i < lim; i++) {
        rez=Perfectnumber(lim);
    }
        cout<<"perfektie skaitļi no 1 lidz "<<lim<<" : "<<endl<<rez<<endl;
    system ("pause");
return 0;
}
//ievadot jau zinamu sk, to ari izvada, bet citus nee

if i have like that is it right, or what i have to correct, because this advice to move cout into loop, dont help.?

Seif · Mar 19th, 2008, 4:50 PM

(Toggle Plain Text)

int Perfectnumber (int rob) {
    int result;
    int sum = 0;
        for (int i = 1; i < rob; i++) {
            if (rob % i == 0) {
               sum += i;
            }
            if (sum == rob) { 
               return true;
            } else {
               return false;       
            }
        }
return true;
}

is wrong for a start

the if sum == rob statement should be outside of the for loop, otherwise your exiting after the first iteration, you need that loop to finish calculating sum before you do this if check.

(Toggle Plain Text)

for (int i = 1; i < lim; i++) {
        rez=Perfectnumber(lim);
    }

I don't understand why you need to check if n is a perfect number n amount of times. Perhaps you should look at your code and understand whats happening.

grumpy · Mar 19th, 2008, 6:41 PM

Quote:

Originally Posted by adenus

if i have like that is it right, or what i have to correct, because this advice to move cout into loop, dont help.?

The advice helps as much as it is possible, given the way you asked your question.

The basic problem in your code is that you have code outside loops that need to be inside the loops, and other things inside loops that need to be outside. Hence the code runs incorrectly. That is enough information for you to fix your problem, if you just think about it a bit.

Mar 19th, 2008, 3:29 PM	#1
adenus Newbie Join Date: Mar 2008 Posts: 3 Rep Power: 0	Why my program dont cout perfect number? (Toggle Plain Text) #include <iostream> using namespace std; //aprekjina funkcija int Perfectnumber (int rob) { int result; int sum = 0; for (int i = 1; i < rob; i++) { if (rob % i == 0) { sum = sum + i; } } if (sum == rob) { result=sum; } return (result); } //galvena funkcija int main() { int rez; int lim; cout<<"ievadi robesu lidz kurai meklet : "; cin>>lim ; for (int i = 1; i < lim; i++) { rez=Perfectnumber(lim); } cout<<"perfect numbers from 1 to"<<lim<<" : "<<rez<<endl; system ("pause"); return 0; } #include <iostream> using namespace std; //aprekjina funkcija int Perfectnumber (int rob) { int result; int sum = 0; for (int i = 1; i < rob; i++) { if (rob % i == 0) { sum = sum + i; } } if (sum == rob) { result=sum; } return (result); } //galvena funkcija int main() { int rez; int lim; cout<<"ievadi robesu lidz kurai meklet : "; cin>>lim ; for (int i = 1; i < lim; i++) { rez=Perfectnumber(lim); } cout<<"perfect numbers from 1 to"<<lim<<" : "<<rez<<endl; system ("pause"); return 0; } why dont it cout perfect numbers, ? and in one forum for me advice to inicialise result, but i dont know which value add to it?

Mar 19th, 2008, 3:59 PM	#2
adenus Newbie Join Date: Mar 2008 Posts: 3 Rep Power: 0	Re: Why my program dont cout perfect number? pls help!

Mar 19th, 2008, 4:11 PM	#4
grumpy Programming Guru Join Date: Jun 2005 Location: Adelaide, South Australia Posts: 1,209 Rep Power: 5	Re: Why my program dont cout perfect number? Don't bump threads. If people have an answer to offer, they will. If you don't get a reply, it means people can't help or are disinclined to. In any event, the code; (Toggle Plain Text) for (int i = 1; i < lim; i++) { rez=Perfectnumber(lim); } cout<<"perfect numbers from 1 to"<<lim<<" : "<<rez<<endl; for (int i = 1; i < lim; i++) { rez=Perfectnumber(lim); } cout<<"perfect numbers from 1 to"<<lim<<" : "<<rez<<endl; only prints out the value of lim, and the value of rez produced in the last iteration through the for loop. If you want to print out every value that rez takes, put the output statement inside the loop.

Mar 19th, 2008, 4:24 PM	#5
adenus Newbie Join Date: Mar 2008 Posts: 3 Rep Power: 0	Re: Why my program dont cout perfect number? (Toggle Plain Text) #include <iostream> using namespace std; //aprekjina funkcija int Perfectnumber (int rob) { int result; int sum = 0; for (int i = 1; i < rob; i++) { if (rob % i == 0) { sum += i; } if (sum == rob) { return true; } else { return false; } } return true; } //galvena funkcija int main() { int rez; int lim; cout<<"ievadi robesu lidz kurai meklet : "; cin>>lim ; for (int i = 1; i < lim; i++) { rez=Perfectnumber(lim); } cout<<"perfektie skaitļi no 1 lidz "<<lim<<" : "<<endl<<rez<<endl; system ("pause"); return 0; } //ievadot jau zinamu sk, to ari izvada, bet citus nee #include <iostream> using namespace std; //aprekjina funkcija int Perfectnumber (int rob) { int result; int sum = 0; for (int i = 1; i < rob; i++) { if (rob % i == 0) { sum += i; } if (sum == rob) { return true; } else { return false; } } return true; } //galvena funkcija int main() { int rez; int lim; cout<<"ievadi robesu lidz kurai meklet : "; cin>>lim ; for (int i = 1; i < lim; i++) { rez=Perfectnumber(lim); } cout<<"perfektie skaitļi no 1 lidz "<<lim<<" : "<<endl<<rez<<endl; system ("pause"); return 0; } //ievadot jau zinamu sk, to ari izvada, bet citus nee if i have like that is it right, or what i have to correct, because this advice to move cout into loop, dont help.?

Mar 19th, 2008, 4:50 PM	#6
Seif Hobbyist Programmer Join Date: Jan 2006 Location: UK Posts: 214 Rep Power: 3	Re: Why my program dont cout perfect number? (Toggle Plain Text) int Perfectnumber (int rob) { int result; int sum = 0; for (int i = 1; i < rob; i++) { if (rob % i == 0) { sum += i; } if (sum == rob) { return true; } else { return false; } } return true; } int Perfectnumber (int rob) { int result; int sum = 0; for (int i = 1; i < rob; i++) { if (rob % i == 0) { sum += i; } if (sum == rob) { return true; } else { return false; } } return true; } is wrong for a start the if sum == rob statement should be outside of the for loop, otherwise your exiting after the first iteration, you need that loop to finish calculating sum before you do this if check. (Toggle Plain Text) for (int i = 1; i < lim; i++) { rez=Perfectnumber(lim); } for (int i = 1; i < lim; i++) { rez=Perfectnumber(lim); } I don't understand why you need to check if n is a perfect number n amount of times. Perhaps you should look at your code and understand whats happening.

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Programming Forums > Application Development > C++

Why my program dont cout perfect number?