1059. Prime Factors (25)

时间限制

50 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：将一个整数分解为若干素数的乘积，注意N==1的情况即可。

#include <iostream>#include <cstdio>#include <cstdlib>#include <algorithm>#include <cstring>#include <cmath>#include <vector>using namespace std;typedef long long int ll;vector< pair<ll,int> > my_vector;int main(){ll num;cin >> num;cout << num << "=";if(num==1){cout << 1 << endl;return 0;}ll tmp = sqrt(num)+1;int count;for(ll i=2; i<=tmp; i++){if(num%i==0){count = 0;do{count++;num /= i;}while(num%i==0);my_vector.push_back({i,count});}}if(num!=1)my_vector.push_back({num,1});int size = my_vector.size();for(int i=0; i<size; i++){cout << my_vector[i].first;if(my_vector[i].second>1)            cout << "^" << my_vector[i].second;        if(i+1!=size)            cout << "*";}    cout <<endl;return 0;}