1059 Prime Factors (25分)
标签：must base vector using for clu size NPU factor
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????.
Each input file contains one test case which gives a positive integer N in the range of long int.
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.
97532468
97532468=2^2*11*17*101*1291
思路：算术基本定理（唯一分解定理）
用从小到大的素数整除n，vector保留素数及其对应的指数
1 #include <iostream> 2 #include <vector> 3 #include <algorithm> 4 5 using namespace std ; 6 7 typedef pair<int,int> PII ; 8 vector<PII> vc ; 9 int n ; 10 11 void get(){ 12 for(int i=2;i<=n/i;i++){ 13 if(n%i==0){ 14 int s = 0 ; 15 while(n%i==0){ 16 s++ ; 17 n /= i ; 18 } 19 vc.push_back({i,s}) ; 20 } 21 } 22 if(n>1){ 23 vc.push_back({n,1}) ; 24 } 25 } 26 27 int main(){ 28 cin >> n ; 29 30 if(n==1){ 31 printf("1=1") ; 32 }else{ 33 printf("%d=",n) ; 34 get() ; 35 int la = vc.size() ; 36 for(int i=0;i<la;i++){ 37 if(i){ 38 printf("*") ; 39 } 40 printf("%d",vc[i].first) ; 41 if(vc[i].second>1){ 42 printf("^%d",vc[i].second) ; 43 } 44 } 45 } 46 47 return 0 ; 48 }
标签：must base vector using for clu size NPU factor
原文地址：https://www.cnblogs.com/gulangyuzzz/p/12034298.html
