1059. Prime Factors (25)

Given any positive integer N, you are supposed to find all of its prime factors, and write them in the format N = p₁^k₁ * p₂^k₂ *…*p_m^k_m.

Input Specification:

Each input file contains one test case which gives a positive integer N in the range of long int.

Output Specification:

Factor N in the format N = p₁^k₁ * p₂^k₂ *…*p_m^k_m, where p_i's are prime factors of N in increasing order, and the exponent k_i is the number of p_i -- hence when there is only one p_i, k_i is 1 and must NOT be printed out.

Sample Input:

97532468

Sample Output:

97532468=2^2*11*17*101*1291

注意n=1的情况！！！

#include <iostream>
#include <map>
#include <queue>
#include <stdio.h>
#include <vector>
#include <algorithm>
#include <stack>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <string>
using namespace std;

#define N 10000
bool mark[N+1];
int prime[N+1];
int primesize;
void init(){
	primesize=0;
	for (int i=2;i<=N;i++)
	{
		if(mark[i]==true)continue;
		prime[primesize++]=i;
		for(int j=i*i;j<=N;j+=i){
			mark[j]=true;
		}
	}
}
int main(){
	init();
	int n;
	int i;
	scanf("%d",&n);
	if (n==1)
	{
		printf("1=1");
		return 0;
	}
	int m=n;
	int ansprime[30];
	int anssize=0;
	int ansnum[30];
	for( i=0;i<primesize;i++){
		if(n%prime[i]==0){
			ansprime[anssize]=prime[i];
			ansnum[anssize]=0;
			while (n%prime[i]==0)
			{
				ansnum[anssize]++;	
				n/=prime[i];
			}
			anssize++;
			if(n==1)break;
		}
	}
	if(n!=1){
		ansprime[anssize]=n;
		ansnum[anssize++]=1;
	}
	int ans=0;
	printf("%d=",m);
	for (i=0;i<anssize;i++)
	{
		if(i==0){
			if(ansnum[i]==1)
				printf("%d",ansprime[i],ansnum[i]);
			else
				printf("%d^%d",ansprime[i],ansnum[i]);
		}else{
			if(ansnum[i]==1)
				printf("*%d",ansprime[i],ansnum[i]);
			else
				printf("*%d^%d",ansprime[i],ansnum[i]);
		}
	}

	
	return 0;
}