Solution of 12895 - Armstrong Number

Problem description 
source: https://uva.onlinejudge.org/external/128/12895.html

Input 

The first line of input is an integer, T that determines the number of test cases. Each of the next T lines contain a positive integer N, where N ≤ 1000000000.


Output

 For each line of input, there will be one line of output. If N is an Armstrong number print ‘Armstrong’, otherwise print ‘Not Armstrong’ (without the quotes).

Sample Input 

3
153
2732
54748

Sample Output 

Armstrong
Not Armstrong
Armstrong

Solution:

#include <algorithm>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <string>
#include<stdio.h>
#include<stdlib.h>
#include<math.h>

using namespace std;

int main() {
    int i, t, n, ArmN, ArmNumbuer, digit;
    while(scanf("%d",&t)==1) {
        for(i=0; i<t; i++) {
            scanf("%d", &ArmN);
            ArmNumbuer = ArmN;
            n = 0;
            while(ArmN > 0) {
                ArmN /= 10;
                n+=1;
            }
            ArmN = ArmNumbuer;
            while(ArmN > 0) {
                digit = ArmN % 10;
                ArmN /= 10;
                ArmNumbuer -= pow(digit, n);
            }
            if(ArmNumbuer == 0)
                printf("Armstrong\n");
            else
                printf("Not Armstrong\n");
        }
    }
    return 0;
}