1059. Prime Factors (25)

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 *…*pm^km.

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:[/b]

97532468

Sample Output:[/b]

97532468=2^2*11*17*101*1291

来源： <http://www.patest.cn/contests/pat-a-practise/1059>

#include <iostream>

#include <vector>

using namespace std;

vector<long int> factor;

vector<long int> exponent;

bool isPrime(long int a) {

if (a == 2 || a == 3)

return true;

for (long int i = 2; i*i < a; i++)

if (a%i == 0)

return false;

return true;

}

int main(void) {

long int n;

cin >> n;

cout << n << "=";

if (n < 2) {

cout << n;

return 0;

}

long int i = 2;

while (true)

{

if (!isPrime(i)) {

i++;

continue;

}

if (n%i == 0) {

int expCnt = 0;

factor.push_back(i);

while (true)

{

if (n%i == 0) {

n = n / i;

expCnt++;

}

else {

exponent.push_back(expCnt);

break;

}

}

}

i++;

if (n == 1) {

break;

}

}

for (int i = 0; i < exponent.size(); i++) {

if (exponent[i] == 1)

cout << factor[i];

else

cout << factor[i] << "^" << exponent[i];

if (i != exponent.size() - 1)

cout << "*";

}

return 0;

}

[/code]

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