1059 Prime Factors (25point(s)) - PAT 甲級 C語言

1059 Prime Factors (25point(s))

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 x … x 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=2^211171011291

題目大意：

給一個整數 N，從小到大輸出該數的質因子

設計思路：

建立一個素數表，遍歷素數表，查找質因子

• INT 最大值爲 2147483647，開根號得 46340.95，素數表大小 5 萬左右
• 所以不能被 5 萬以內整除的數一定是素數
• 當最後 N > 1 時，state 是爲了標示是否輸出過因子，例如 30027 = 3 * 10009，最後輸出 10009 前面要有 * 號

編譯器：C (gcc)

#include<stdio.h>

#define MAX 46342
//INT 2147483647 46340.95
int main(void) {
        int prime[MAX] = {-1, -1};
        int i, j;
        for (i = 2; i * i < MAX; i++) {
                for (j = 2; i * j < MAX; j++) {
                        prime[i * j] = -1;
                }
        }

        int n;
        scanf("%d", &n);
        printf("%d=", n);
        if (n == 1) {
                printf("1");
        }
        int state = 0;
        for (i = 2; i < MAX && n >= 2; i++) {
                int count = 0, flag = 0;
                while (prime[i] == 0 && n % i == 0) {
                        count++;
                        n /= i;
                        flag = 1;
                }
                if (flag) {
                        if (state) {
                                printf("*");
                        }
                        printf("%d", i);
                        state = 1;
                }
                if (count > 1) {
                        printf("^%d", count);
                }
        }
        if (n > 1) {
                printf("%s%d", state ? "*": "", n);
        }

        return 0;
}