CODEWITHSUNDEEP
c
angellee0102
angellee0102

Reputation: 61

Armstrong Numbers between 1-10000 using C

I'm trying to print out Armstrong Numbers from 1-10000, and my question is that there are "four 1s" when i compile and run, and I would like to ask which part of coding that I did wrong, and what part of code should be revised if I want only one 1 to be printed out. Other than that, all the others Armstrong Numbers output correctly. It's my first time asking here and I hope people who sees this would give me some advice. Thank you.

#include <stdio.h>
#include <stdlib.h>

/* run this program using the console pauser or add your own getch, system("pause") or input loop */

int main(int argc, char *argv[]) {
printf("Armstrong Numbers from 1-10000:\n");
int digit1, digit2, digit3, digit4; 
int i;

for(i=1; i<10000; i++){

    digit4=i/1000;
    digit3=(i%1000)/100;
    digit2=((i%1000)%100)/10;
    digit1=((i%1000)%100)%10;

    //one digit number 
    if(i<10){
        if(i==digit1)printf("%d\n",i);
    }
    //two digit number
    if(10<=i<100){
        int output100 = digit1*digit1 + digit2*digit2;
        if(i==output100)printf("%d\n",i);
    }
    //three digit number 
    if(100<=i<=999){
        int output1000 = digit1*digit1*digit1 + digit2*digit2*digit2 + digit3*digit3*digit3;
        if(i==output1000){
            printf("%d\n",i);
        }
    }
    //four digit number
    if(1000<=i<=10000){
        int output10000 = digit1*digit1*digit1*digit1 + digit2*digit2*digit2*digit2 + digit3*digit3*digit3*digit3 + digit4*digit4*digit4*digit4;
        if(i==output10000){
            printf("%d\n",i);
        }
    }
    }


return 0;
}

Upvotes: 0

Views: 2012

Answers (1)

David Ranieri
David Ranieri

Reputation: 41027

Compile with warnings:

if(1000<=i<=10000){

warning: comparisons like ‘X<=Y<=Z’ do not have their mathematical meaning [-Wparentheses]

Do you mean if(i >= 1000 && i < 10000){ ?

Upvotes: 8

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