【笨方法學PAT】1059 Prime Factors (25 分)
阿新 • • 發佈:2018-12-20
一、題目
Given any positive integer N, you are supposed to find all of its prime factors, and write them in the format N = p1k1×p2k2×⋯×pmkm.
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
Sample Input:
97532468
Sample Output:
97532468=2^2*11*17*101*1291
二、題目大意
因式分解。
三、考點
素數表
四、注意
1、使用笨方法建立素數表,嚴重超時,記住這個建立素數表的方法;
2、參考:https://www.liuchuo.net/archives/2289。
五、程式碼
#include <cstdio> #include <vector> using namespace std; vector<int> prime(500000, 1); int main() { for(int i = 2; i * i < 500000; i++) for(int j = 2; j * i < 500000; j++) prime[j * i] = 0; long int a; scanf("%ld", &a); printf("%ld=", a); if(a == 1) printf("1"); bool state = false; for(int i = 2; a >= 2;i++) { int cnt = 0, flag = 0; while(prime[i] == 1 && a % i == 0) { cnt++; a = a / i; flag = 1; } if(flag) { if(state) printf("*"); printf("%d", i); state = true; } if(cnt >= 2) printf("^%d", cnt); } return 0; }