PAT Advanced 1059 Prime Factors (25) [素数表的建⽴]

题目

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 …pm^km.

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 …pm^km, where pi’s are prime factors of N in increasing order, and the exponent ki is the number of pi — hence when there is only one pi, ki is 1 and must NOT be printed out.

Sample Input:

97532468

Sample Output:

97532468=2^211171011291

题目分析

输入一个整数N，将其分解为质因数的乘法式

解题思路

1. 创建质数表
2. 依次判断质数表中质数i，若N%i==0，则打印质数（并内循环统计i整除N的次数tm，每次内循环完成更新N=N/i，若N%i!=0，退出内循环）;
3. 打印当前质数的指数tm

Code

Code 01

#include <iostream>
#include <cmath>
using namespace std;
bool isPrime(int n) {
	if(n<=1)return false;
	int sqr=(int)sqrt(1.0*n);
	for(int i=2; i<=sqr; i++) {
		if(n%i==0)return false;
	}
	return true;
}
int prime_table[100010];
bool prime_bool[100010];
void create_pt() {
	int index = 0;
	for(int i=2; i<100010; i++) {
		if(prime_bool[i]==false) {
			prime_table[index++]=i;
			for(int j=i+i; j<100010; j+=i) {
				prime_bool[j]=true;
			}
		}
	}
}
int main(int argc,char *argv[]) {
	// 1.创建素数表
	create_pt();
	// 2.接收输入
	long long n;
	scanf("%lld", &n);
	// 3. 打印
	printf("%lld=",n);
	if(n==1)printf("1");
	int index=0;
	bool wf = false; //false为打印的第一个数字，true为打印的第一个数字后的数字
	while(n>1) {
		int prime = prime_table[index];
		bool flag = false;
		int tm=0;
		while(n%prime==0) {
			n/=prime;
			tm++;
		}
		if(tm>0) {
			if(wf)printf("*");
			printf("%d",prime);
			wf = true;
		}
		if(tm>1)printf("^%d",tm);
		index++;
	}
	return 0;
}

Code 02（素数表生成简洁但不高效）

#include <cstdio>
#include <vector>
using namespace std;
vector<int> prime(500000, 1);
int main() {
	for(int i = 2; i * i < 500000; i++)
		for(int j = 2; j * i < 500000; j++)
			prime[j * i] = 0;
	long int a;
	scanf("%ld", &a);
	printf("%ld=", a);
	if(a == 1) printf("1");
	bool state = false;
	for(int i = 2; a >= 2; i++) {
		int cnt = 0, flag = 0;
		while(prime[i] == 1 && a % i == 0) {
			cnt++;
			a = a / i;
			flag = 1;
		}
		if(flag) {
			if(state) printf("*");
			printf("%d", i);
			state = true;
		}
		if(cnt >= 2)
			printf("^%d", cnt);
	}
	return 0;
}