1096 Consecutive Factors (20point(s)) - C語言 PAT 甲級

1096 Consecutive Factors (20point(s))

Among all the factors of a positive integer N, there may exist several consecutive numbers. For example, 630 can be factored as 3×5×6×7, where 5, 6, and 7 are the three consecutive numbers. Now given any positive N, you are supposed to find the maximum number of consecutive factors, and list the smallest sequence of the consecutive factors.

Input Specification:

Each input file contains one test case, which gives the integer N (1<N<2³¹).

Output Specification:

For each test case, print in the first line the maximum number of consecutive factors. Then in the second line, print the smallest sequence of the consecutive factors in the format factor[1]factor[2]…*factor[k], where the factors are listed in increasing order, and 1 is NOT included.

Sample Input:

630

Sample Output:

3
567

題目大意：

輸入整數 N，求 N 的最小的連續因子序列

設計思路：

直接暴力循環，枚舉以 2 ~ 根號 N 開頭的每個序列，最後序列長度爲 0，說明是質數

編譯器：C (gcc)

#include <stdio.h>
#include <math.h>

int main(void)
{
        int n, m;
        int start = 0, cnt = 0;
        int i, j, k;

        scanf("%d", &n);
        m = (int)sqrt(n);
        for (i = 2; i <= m; i++) {
                k = 0;
                for (j = i; n % j == 0; j *= (i + k))
                        k++;
                if (k > cnt) {
                        cnt = k;
                        start = i;
                }
        }
        if (cnt == 0) {
                printf("1\n%d", n);
        } else {
                printf("%d\n%d", cnt, start);
                for (i = 1; i < cnt; i++)
                        printf("*%d", start + i);
        }
        return 0;
}