1059 Prime Factors (25分)

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本文介绍了一种质因数分解算法，通过筛选法找出所有质数，然后将输入的正整数分解为其质因数的乘积形式，并以特定格式输出。例如，将97532468分解为2^2*11*17*101*1291。

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

题意：质因数分解，把一个大数分解成多个质数乘积，若一个数出现k次，那么就用这个质数的k次方表示。

#include<iostream>
#include<algorithm>
using namespace std;
int main(){

    int is_prime[40000];
    
    fill(is_prime, is_prime + 40000, 1);
    is_prime[0] = is_prime[1] = 0;
    for(int i = 2; i * i < 40000; i++){
        for(int j = 2; j * i < 40000; j++){
            is_prime[i * j] = 0;
        }
    }
    int n;
    scanf("%d", &n);
    printf("%d=", n);
    if(n == 1){
        printf("1\n");
        return 0;
    }
    for(int i = 2; n >1; i++){
        int cnt = 0;
        while (is_prime[i] && n % i == 0){
            cnt++;
            n = n / i;
            if(cnt == 1){
                printf("%d", i);
            }
        }
        if(cnt > 1){
            printf("^%d", cnt);
        }
        if(cnt && n > 1){
            printf("*");
        }
        

    }
    return 0;
}