1059 Prime Factors(25 分)发布时间:2022/5/31 12:58:53
Given any positive integer N, you are supposed to find all of its prime factors, and write them in the format N = p1k1×p2k2×⋯×pmkm.
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*…*pm^km, where pi's are prime factors of N in increasing order, and the exponent ki is the number of pi -- hence when there is only one pi, ki is 1 and must NOT be printed out.
Sample Input:
97532468
Sample Output:
97532468=2^2*11*17*101*1291
#include
#include
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 ){
if(n % i == 0) return false;
}
return true;
}
int prime[maxn],pNum = 0;
void Find_prime(){
for(int i = 1 ; 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 ){
//printf("%d",i);
if(n % prime == 0){
fac[num].x = prime;
fac[num].cnt = 0;
while(n % prime == 0){
fac[num].cnt++;
n /= prime;
}
num++;
}
if(n == 1) break;
}
if(n != 1){
fac[num].x = n;
fac[num].cnt = 1;
}
//printf("1\n");
for(int i = 0; i ){
if(i > 0) printf("*");
printf("%d",fac.x);
if(fac.cnt > 1) printf("^%d",fac.cnt);
}
}
return 0;
}
转载于:https://www.cnblogs.com/wanghao-boke/p/9532827.html

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