PAT (Advanced Level) Practice 1059 Prime Factors

                                          1059 Prime Factors

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​​​​.

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 = 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.

Sample Input:

97532468

Sample Output:

97532468=2^2*11*17*101*1291

实现代码：

#include<cstdio>
#include<cmath>
const int maxn = 100010;
bool is_prime(int n)
{
	if(n == 1) return false;
	int sqr = (int) sqrt(1.0*n);
	for(int i=2;i<=sqr;i++){
		if(n%i==0) return false;
	}
	return true;
 } 
 int prime[maxn],pNum=0;
 void Find_Prime(){
 	for(int i=1;i<maxn;i++ ){
 		if(is_prime(i)==true){
 			prime[pNum++] = i;
		 }
	 }
 }
 struct factor{
 	int x,cnt;
 	
 }fac[10];
 int main()
 {
 	Find_Prime();
 	int n,num=0;
 	scanf("%d",&n);
 	if(n == 1) printf("1=1");
 	else{
 		printf("%d=",n);
 		int sqr = (int) sqrt(1.0*n);
 		for(int i=0;i<pNum&&prime[i]<=sqr;i++){
 			if(n%prime[i]==0){
 				fac[num].x=prime[i];
 				fac[num].cnt=0;
 				while(n%prime[i]==0){
 					fac[num].cnt++;
 					n/=prime[i];
				 }
				 num++;
			 }
			 if(n == 1) break;
			 
		 }
		 if(n != 1){
		 	fac[num].x=n;
		 	fac[num++].cnt=1;
		 }
		 for(int i=0;i<num;i++){
		 	if(i>0) printf("*");
		 	printf("%d",fac[i].x);
		 	if(fac[i].cnt>1){
		 		printf("^%d",fac[i].cnt);
			 }	 
		}
	 }
	 return 0;
 }

 

结果如下：