PAT A1059. Prime Factors (25)

1059. Prime Factors (25)

时间限制

100 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

思路分析：

打印从1到sqrt(n)的素数表，然后存在prime中，挨个确认是不是n的质因数并确定其个数。最后确定是否有大于sqrt(n)的质因数。

题解：

#include <cstdio>#include <algorithm>#include <cmath>using namespace std;const int MAX = 100000;int n, range, prime[MAX] = {0}, num = 0;bool isprime[MAX] = {false};//表示是素数 struct Factor{int x, cnt;}fac[20];void print_prime(){isprime[1] = true;range = floor(sqrt(n));for(int i = 2; i <= range; i++){if(!isprime[i]){for(int j = i + i; j <= range; j += i){isprime[j] = true;}}}}int main(){scanf("%d", &n);int m = n;if(n == 1){printf("1=1");return 0;} print_prime();for(int i = 1; i <= range; i++){if(isprime[i] == false){prime[num++] = i;}}int fnum = 0;for(int i = 0; i < num; i++){if(n % prime[i] == 0){fac[fnum].x = prime[i];fac[fnum].cnt = 0;while(n % prime[i] == 0){n /= prime[i];fac[fnum].cnt++;}fnum++;}if(n == 1) break;}if(n > 1){fac[fnum].x = n;fac[fnum++].cnt = 1;}printf("%d=", m);for(int i = 0; i < fnum; i++){printf("%d", fac[i].x);if(fac[i].cnt > 1) printf("^%d", fac[i].cnt);if(i != fnum-1) printf("*");}return 0;}