PAT (Advanced Level) Practise 1059 Prime Factors (25)

最新推荐文章于 2017-08-09 19:21:31 发布

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最新推荐文章于 2017-08-09 19:21:31 发布

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本文介绍了一种质因数分解算法，通过预先筛选出素数并遍历这些素数来分解给定的正整数。文章提供了一个完整的C++代码实现，并通过样例输入输出展示了算法的具体应用。

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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

题意：给你一个数，将它拆成几个素数的乘积

解题思路：素数筛找出所有素数，然后暴力找出哪些素数能被整除

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <functional>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;

int visit[1000009], prime[1000009];
LL n;

void init()
{
	memset(visit, 1, sizeof visit);
	visit[0] = visit[1] = 0;
	int cnt = 0;
	for (int i = 2; i <= 1000005; i++)
	{
		if (!visit[i]) continue;
		prime[cnt++] = i;
		for (int j = i * 2; j <= 1000005; j += i) visit[j] = 0;
	}
}

int main()
{
	init();
	while (~scanf("%lld", &n))
	{
		printf("%d=", n);
		if (n == 1) { printf("%d", n); continue; }
		int flag = 0;
		for (int i = 0; prime[i] <= n; i++)
		{
			if (n%prime[i]) continue;
			int sum = 0;
			while (n%prime[i] == 0) { sum++; n /= prime[i]; }
			if (flag) printf("*");
			flag = 1;
			printf("%d", prime[i]);
			if (sum > 1) printf("^%d", sum);
		}
		printf("\n");
	}
	return 0;
}