1059 Prime Factors（25 分)(cj)

2024-06-20 02:19:16

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

思路, 一个坑就是如果 n无法被 $\sqrt{n}$ 之内的素数整除，那么n就是素数本身，打印本身。

筛选素数用埃拉托色尼（Eratosthenes）筛法就可以。

code

#pragma warning(disable:4996)
#include <iostream>
#include <vector>
#include <iostream>
#include <math.h>
using namespace std;
void init(int n);
vector<int> varr;
int main(){long long n;cin >> n;cout << n << '=';int maxn = sqrt(n);init(maxn);bool f = 1;for (int i = 0; i < varr.size(); ++i) {int a = varr[i];if (a > n) break;int b = 0;while (n%a == 0) {++b;n /= a;}if (b > 0) {if (f) f = 0;else cout << '*';if (b == 1) cout << a;else cout << a << '^' << b;}}if (f) cout << n << endl;system("pause");return 0;
}
void init(int n) {int* arr = new int[n+1];for (int i = 0; i <= n; ++i) arr[i] = 1;arr[1] = 0;for (int i = 2; i <= n; ++i) {if (arr[i] == 1) {varr.push_back(i);int p = i*2;while (p <= n) {arr[p] = 0;p += i;}}}
}

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