PAT 甲级 1059 Prime Factors（25 分）_pat甲级1059-优快云博客

本文链接：https://blog.youkuaiyun.com/Hello_Mr_X/article/details/81986412

本文介绍了一种质因数分解算法，用于将任意正整数分解为其所有质因数的乘积形式。输入为一个正整数N，算法将输出N的所有质因数及其对应的指数，格式为N = p1^k1 * p2^k2 * ... * pm^km，其中pi为N的质因数，ki为pi的次数。

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1059 Prime Factors（25 分）

Given any positive integer N, you are supposed to find all of its prime factors, and write them in the format N = p1k1×p2k2×⋯×pmkm.

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

# include <iostream>
# include <cstdio>
# include <cmath>
using namespace std;
const int MAXN = 1e5;
int plist[MAXN], pnum = 0;

struct zyz{
	int p, c;
}prime[10];

bool is_prime(int n){
	if(n == 1) return false;
	int sqr = sqrt(1.0 * n);
	for(int i = 2; i < sqr; i++){
		if(n % i == 0)
			return false;
		else return true;
	}
}

void Find_prime(){
	for(int i = 2; i < MAXN; i++){
		if(is_prime(i))
			plist[pnum++] = i;
	}
}

int main(){
//	freopen("C:\\1.txt", "r", stdin);
	Find_prime();
	int n, num = 0;
	scanf("%d", &n);
	if(n == 1){
		printf("1=1");
		return 0;
	}
	printf("%d=", n); 
	int sqr = sqrt(1.0 * n);
	for(int i = 0; i < pnum && plist[i] < sqr; i++){
		if(n % plist[i] == 0){
			prime[num].p = plist[i];
			prime[num].c = 0;
			while(n % plist[i] == 0){
				prime[num].c++;
				n /= plist[i];
			}
			num++;
		}
		if(n == 1) break;
	}
	if(n != 1){
		prime[num].p = n;
		prime[num++].c = 1;
	}
	
	for(int i = 0; i < num; i++){
		if(i > 0) printf("*");
		printf("%d", prime[i].p);
		if(prime[i].c > 1)
			printf("^%d", prime[i].c);
	}
	return 0;
}