PAT甲级练习1059. Prime Factors (25)

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最新推荐文章于 2022-12-02 22:14:11 发布

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本文介绍了一种求解并输出正整数素因数分解的方法。通过从2开始逐个检查，直到找到所有素因数及其指数，并按格式输出。示例中使用了97532468作为输入。

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

一道求素数因数的题，直接从2开始强解即可

#include <iostream>
#include <cstdio>
#include <math.h>
#include <algorithm>
#include <vector>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <string>
#include <string.h>

using namespace std;

const int MAX=1e5+10;

int main(){
	long n, flag=0;
	scanf("%ld", &n);
	printf("%ld=", n);
	vector<long> f;

	for(int i=2,j; i*i<=n; i++){
		for(j=0; n%i==0; n/=i) j++;
		if(j){
			if(flag) printf("*"); else flag=1;
			if(j>1) printf("%ld^%ld", i, j); else printf("%ld",i);  
		}
	}
	if(flag){
		if(n>1) printf("*%ld\n",n);
	}else
		printf("%ld\n",n);

	cin>>n;
	return 0;
}