Algorithm
problem link- https://www.spoj.com/problems/PTIME/
PTIME - Prime Time
For your math homework this week your teacher gave you five large numbers and asked you to find their prime factors. However these numbers aren't nearly large enough for someone with knowledge of programming like yourself. So you decide to take the factorial of each of these numbers. Recall that N! (N factorial) is the product of the integers from 1 through N (inclusive). It’s your job now to create a program to help you do your homework.
Input
Each test case contains a number N (2 ≤ N ≤ 10000).
Output
The output should contain a line representing the prime factorization of the factorial given number, which should be of the form: p1^e1 * p2^e2 * ... * pk^ek where p1, p2, ... pk are the distinct prime factors of the factorial of the given number in increasing order, and e1, e2, ... ek are their exponents.
Example
Input: 10 Output: 2^8 * 3^4 * 5^2 * 7^1
Code Examples
#1 Code Example with C++ Programming
Code -
C++ Programming
#include <bits/stdc++.h7gt;
#include <limits.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> ii;
typedef vector<ii> vii;
typedef vector<int> vi;
#define MOD (ll)1000000007
#define pb push_back
#define EPS 1e-9
#define FOR(i, n) for(int i = 0;i < n; i++)
#define pi(a) printf("%d\n", a)
#define all(c) c.begin(), c.end()
#define tr(container, it) for(typeof(container.begin()) it = container.begin(); it != container.end(); it++)
#define gc getchar_unlocked
template <typename T> T gcd(T a, T b){return (b==0)?a:gcd(b,a%b);}
template <typename T> T lcm(T a, T b){return a*(b/gcd(a,b));}
template <typename T> T mod_exp(T b, T p, T m){T x = 1;while(p){if(p&1)x=(x*b)%m;b=(b*b)%m;p=p>>1;}return x;}
template <typename T> T invFermat(T a, T p){return mod_exp(a, p-2, p);}
template <typename T> T exp(T b, T p){T x = 1;while(p){if(p&1)x=(x*b);b=(b*b);p=p>>1;}return x;}
void si(int &x){
register int c = gc();
x = 0;
int neg = 0;
for(;((c<48 || c>57) && c != '-');c = gc());
if(c=='-') {neg=1;c=gc();}
for(;c>47 && c<58;c = gc()) {x = (x<<1) + (x<<3) + c - 48;}
if(neg) x=-x;
}
int factors[10005];
void div(int n){
for(int i = 2;i*i <= n; i++){
if(n%i == 0){
while(n%i == 0){
factors[i]++;
n/=i;
}
}
}
if(n!=1){
factors[n]++;
}
}
int main(){
// io;
int n;
cin>>n;
for(int i = 2;i <= n; i++){
div(i);
}
for(int i = 2;i <= 10005; i++>{
if(factors[i] > 0){
cout<<i<<"^"<<factors[i]<<" ";
i++;
while(i <= 10005){
if(factors[i] > 0){
cout<<"*"<<" ";
i--;
break;
}
i++;
}
}
}
return 0;
}Input
Output
Demonstration
SPOJ Solution-Prime Time-Solution in C, C++, Java, Python