uva-160-Factors and Factorials solve with cpp / c++

 uva-160-Factors and Factorials solve with cpp / c++
............................................................................................

#include <iostream> #include <bits/stdc++.h> using namespace std; int main () { long n, i, flag, count, j, save; long b [] = {2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97}; while (scanf ("%ld", &n)) { if (n == 0) return 0; save = n; long a [25] = {0}; while (n != 1) { i = n; j = 0; while (i > 1) { count = 0; while (i % b [j] == 0) { i /= b [j]; count++; } a [j] += count; j++; } n--; } // output for (i = 24; i >= 0; i--) { if (a [i] != 0) break; } j = i; printf("%3ld! =", save); if (j >= 15) flag = 0; else flag = 15; for (i = 0; i <= j; i++) { printf("%3ld", a [i]); flag++; if (flag == 15) printf("\n%6c",' '); } printf("\n"); } return 0; } /* N! for N = 100 will not fit into any integral type, you must determine divisibility in another manner. And also be cautious that you are not printing a newline when factors size in 15. Calculate the factorial of given number n and then find the prime factor of factorial n (n!) for example: 5 ! = 120 = 2 * 2 * 2 * 3 * 5 6 ! = 720 = 2 * 2 * 2 * 2 * 3 * 3 * 5 this destructive process should not be implemented .. rather, we can find a far better way to do that, lets see 5 ! = 120 = 5 * 4 * 3 * 2 write a procedure that can count prime factors of a given number prime_factor (5) + prime_factor (4) + prime_factor (3) + prime_factor (2) = prime_factor (5!) = prime_factor (120) similarly, 6 ! = 720 = 6 * 5 * 4 * 3 * 2 prime_factor (6) + prime_factor (5) + prime_factor (4) + prime_factor (3) + prime_factor (2) = prime_factor (6!) = prime_factor (720) prime_factor of (N) continue while (N > 1) { continue while (N % prime = 0) N /= prime; prime = Next_prime (); } // u need to calculate just these prime factors, 2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53, 59,61,67,71,73,79,83,89,97 // u must not print such prime factors position which frequency is zero output format: 1. Number N have to be printed right justified of width 3 2. Each of occurrence will be printed again, right justified of field width 3 3. if there r more than 15 occurrence in one line then u have to print a newline and start to print next occurrence with a field width 6 */