SOJ2666 Decomposition of Problem solving report n!

SOJ2666 decomposition of Problem solving report n!

Description give you a number n (1 < n <= 1000000), Beg n! The factorization form of the mass factor (n factorial) is n=p1^m1*p2^m2*p3^m3......*  here P1 < P2 < P3 < ... For prime numbers *  if mi = 1, then the   ^ mi   input input is multiple case, one number per line n,1 < n <= 1000000, when n equals 0 o'clock input end   output one row per n for its mass factorization form, sample Inputsample output   Author  windy7926778

The main idea of the topic: slightly.
Analysis: The feeling is the bare prime number Operation property. There is a property of prime number, in the decomposition of n!, the quantity of themass factor P is pn (n,p) =[n/p]+[n/p^2]+[n/p^3]+ ... Until [N/p^q]==0, where [] means rounding down. So we're going to break down the n!. You can first use the number of sieve to hit the table, and then decomposition by one to see how many, and finally output results can be. Note The format of the output.
on the code:

#include <iostream> #include <cmath> #include <cstring>using namespace Std;int isprime[1010000];long  Long Prime[1010000];int pnum[1010000];int cnt;void Getp () {for (int i = 1; i < 1010000; i++) Isprime[i] = 1;for (int i = 2; i < 1010000; i++) {if (!isprime[i]) continue;prime[cnt++] = i;for (int j = 2 * i; J < 1010000; J + = i) isprime[j] = 0;}} Long long get (long long n, long long key) {int tem = Key;long Long num = 0;while (n >= key) {num + = N/key;key *= tem;} return num;} int main () {Long long n;getp (), while (CIN >> n &&n) {if (n = = 1) {cout << "1=1" << endl;continue;} memset (pnum, 0, sizeof pnum); for (int i = 0; Prime[i] <= n&&prime[i]; i++) {Pnum[i] = Get (n, Prime[i]);} cout << n << "="; bool FST = 0;for (int i = 0; Prime[i] <= n&&prime[i]; i++) {if (!fst) FST = True;else cout << "*"; if (Pnum[i]) cout << prime[i];if (pnum[i]>=2) cout << "^" << pnum[i];} cout << Endl;} return 0;}

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SOJ2666 Decomposition of Problem solving report n!