1059. Prime Factors

1059. Prime Factors (25)

时间限制

50 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

#include<stdio.h>using namespace std;bool flag[100002];int prime[100002];int primesize;void init(){primesize = 0;for(int i = 2; i <= 100000; i ++){if(flag[i])continue;prime[primesize ++] = i;if(i >= 1000)continue;for(int j = i*i; j <= 100000; j += i)flag[j] = true;}}int main(){init();int n, m;scanf("%d", &n);m = n;if(m == 1){printf("1=1\n");return 0;}int ansPrime[30], ansNum[30];int ansSize = 0;for(int i = 0; i < primesize; i ++){if(m % prime[i] == 0){ansPrime[ansSize] = prime[i];ansNum[ansSize] = 0;while(m % prime[i] == 0){ansNum[ansSize] ++;m /= prime[i];}ansSize ++;if(m == 1)break;}}if(m != 1){ansPrime[ansSize] = m;ansNum[ansSize] = 1;}printf("%d=", n);for(int i = 0; i < ansSize; i ++){if(i == ansSize-1){if(ansNum[i] == 1)printf("%d\n", ansPrime[i]);elseprintf("%d^%d\n", ansPrime[i], ansNum[i]);}else{if(ansNum[i] == 1)printf("%d*", ansPrime[i]);elseprintf("%d^%d*", ansPrime[i], ansNum[i]);}}return 0;}