1059 Prime Factors (25 分)

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 = p​1​​​k​1​​​​×p​2​​​k​2​​​​×⋯×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 = p​1​​^k​1​​*p​2​​^k​2​​*…*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素数筛，可能也有点超纲。。。但是也不难。 1 #include <bits/stdc++.h> 2 #define mod 2000005 3 #define ll long long 4 using namespace std; 5 int an[mod], prime[mod]; 6 vector<int> v,vt; 7 int pos = 0; 8 void init(){ 9 an[0] = an[1] = 1; 10 for(int i=2; i < mod; i++){ 11 if(an[i] == 0){ 12 prime[pos++] = i; 13 for(int j = 2; j*i < mod; j++) 14 an[i*j] = 1; 15 } 16 } 17 } 18 19 ll n; 20 21 int main(){ 22 init(); 23 24 cin >> n; 25 if(n == 1){ 26 cout <<n<<"="<<n<<endl; 27 return 0; 28 } 29 cout <<n<<"="; 30 for(int i = 0; i < pos; i++){ 31 if(n%prime[i] == 0){ 32 int count = 0; 33 while(n%prime[i] == 0){ 34 n /= prime[i]; 35 count++; 36 if(n == 0) 37 break; 38 } 39 v.push_back(prime[i]); 40 vt.push_back(count); 41 } 42 if(n == 1){ 43 break; 44 } 45 } 46 for(int i = 0; i < v.size(); i++){ 47 if(vt[i] == 1){ 48 cout << v[i]; 49 }else{ 50 cout << v[i] <<"^"<<vt[i]; 51 } 52 printf("%c",i == v.size()-1?'\n':'*'); 53 } 54 55 return 0; 56 }

 

转载于:https://www.cnblogs.com/zllwxm123/p/11190910.html

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