PAT A1059 质因子分解

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1059. Prime Factors (25)

时间限制

100 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

HE, Qinming

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

大意是将一个long int型整数进行质因子分解

解题思路：

1.打质数表

2.将所给整数m按照质数表从小到大的顺序相除，每个质数都除到不能除为止，用结构体记录相应质数和除的次数

3.输出

坑：

1.等于1的时候是特例，得特殊处理

2.注意最后得到的1不能输出

代码（long long和int用混了，不管了反正ac了）

#include<cmath>
#include<time.h>
#include<cstdlib>
#include<cstdio>
#include<cstring>
using namespace std;

typedef long long LL;
#define MAX 100000000

LL prime[10000];
int count = 0;
bool is_prime[MAX] = {0};

void get_prime(int maxn){
	memset(is_prime,0,sizeof(bool)*maxn);
	for(int i=2;i<maxn;i++){
		if(!is_prime[i]){
			prime[count++] = i;
			for(int j=i+i;j<maxn;j += i){
				is_prime[j] = true;
			}
		}
	}
}

int main(){
	LL mm,m;
	scanf("%lld",&mm);
	if(mm == 1){
		printf("1=1");
		return 0;
	}
	m = mm;
	LL m_sqrt = sqrt(m*1.0);
	get_prime(m_sqrt);

	struct ans{
		LL prime;
		int num;
	}ans[10];
	
	int ans_c = 0;
	LL i=0;
	while(i<count){
		if(m%prime[i]==0){
			ans[ans_c].prime = prime[i];
			ans[ans_c].num = 1;
			m = m/prime[i];
			while(m%prime[i]==0){
				m = m/prime[i];
				ans[ans_c].num++;
			}
			ans_c++;
		}
		i++;
	}
	if(m!=1){
		ans[ans_c].prime = m;
		ans[ans_c++].num = 1;
	}
	printf("%lld=",mm);
	for(i=0;i<ans_c;i++){
		if(i>0) printf("*");
		if(ans[i].num!=0){
			if(ans[i].num == 1){
				printf("%lld",ans[i].prime);
			}else{
				printf("%lld^%d",ans[i].prime,ans[i].num);
			}
		}
	}
}