題目
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:
97532468
Sample Output:
97532468=2^2*11*17*101*1291
思路
爲了避免重複運算,維護一個素數表prime,初始化爲{2}。
分解N時,先考察素數表中的數是否是其因子;若素數表中沒有其因子,且素數表中最大值比根N小,那麼逐個考察到根N,每發現一個素數都保存到素數表中。
把N的因子m保存到factor數組中,再迭代考察n / m;若其沒有因子,則其本身就是素數,將其保存到factor中。
最後按指定規則輸出結果即可。
注意輸入的值可能是1,需單獨考慮,否則測試點3無法通過。
代碼
#include <iostream>
#include <cmath>
#include <vector>
using namespace std;
vector<int> prime; //素數備忘錄
vector<int> factor; //素因數
bool isPrime(int n){
for (int i=2; i<=sqrt(n); i++){
if (n % i == 0){
return false;
}
}
return true;
}
void divide(int n){
if (n < 2){
return;
}
int len = prime.size();
int m = 0;
for (int i=0; i<len; i++){
if (n % prime[i] == 0) {
m = n / prime[i];
factor.push_back(prime[i]);
break;
}
}
if (m==0){ //n不能被備忘錄中的素數整除,繼續考察根N範圍內是否有因數,擴展素數備忘錄
for (int i=prime[len-1]+1; i<=sqrt(n); i++){
if (isPrime(i)){
prime.push_back(i);
if (n % i == 0){
m = n / i;
factor.push_back(i);
break;
}
}
}
}
if (m==0){ //n確定是素數
factor.push_back(n);
return;
}
else {
divide(m);
}
}
int main(){
prime.push_back(2);
int n;
cin >> n;
cout << n << "=";
if (n==1){
cout << "1";
}
else{
divide(n);
int i = 0;
while (i<factor.size()){
if (i > 0){
cout << "*";
}
cout << factor[i];
int j = i + 1;
while (j<factor.size() && factor[j]==factor[i]){
j++;
}
int k = j - i;
if (k > 1){
cout << "^" << k;
}
i = j;
}
}
return 0;
}