pat 1059. Prime Factors

原题链接：https://www.patest.cn/contests/pat-a-practise/1059

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

题意：找一个数的素因子，按照从小到大的顺序输出。。

解：挨着找就是了，注意特判n=1的情况，因为这个WA了两次。

代码：

#include <cstdio>#include <cstring>#include <cmath>typedef long long ll;int isprime(int x){int i, t = sqrt(x);if (x == 0 || x == 1)return 0;for (i = 2; i <= t; i++){if (x%i == 0)return 0;}return 1;}int main(){int i, j = 1;ll n;scanf("%lld", &n);int cnt = 0;printf("%lld=", n);if (n == 1){printf("1\n");return 0;}for (i = 2; i <= n; i++){if (isprime(i) && n%i == 0){if (j > 1)printf("*");while (n % i == 0){cnt++;n /= i;}printf("%d", i);if (cnt > 1)printf("^%d", cnt);if (n == 1)break;j++;cnt = 0;}}printf("\n");return 0;}