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 #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 facot{int x,cnt; }fac[10]; int main(){Find_prime();int n;scanf("%d",&n);int num = 0;if(n == 1) printf("1=1");else{printf("%d=",n);int sqr = (int)sqrt(1.0*n);//printf("prime[0]");for(int i = 0; i < pNum ; i++){//printf("%d",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;}//printf("1\n");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; }

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轉載于:https://www.cnblogs.com/wanghao-boke/p/9532827.html

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