PAT(A) - 1096. Consecutive Factors (20)

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本文介绍了一种算法，用于找出一个正整数的所有因子中，最长的连续因子序列，并给出具体的实现代码。输入一个正整数N，输出最长连续因子的数量及最小的连续因子序列。

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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:

Sample Output:

3
5*6*7

思路分析：一个分解因子的问题，要去找出最长的连续因子。我的做法是从2开始遍历（题目说1除外），依次判断当前数组能否被N整除，边找边记录连续因子的长度，如果整除不尽，找到第一个因子的位置，我的程序里是head = j - curLen + 1，j为当前整除不尽的因子，curLen为当前连续因子的长度，所以head就出来了。最后输出连续因子的长度。如果最后这个数只能被自身整除，那就输出它自己。

#include <cstdio>
#include <cmath>

int main() {
    int n;
    scanf( "%d", &n );

    int sqr = (int)sqrt( n );
    int maxLen = 0;
    int head = 0;

    for( int i = 2; i <= sqr; i++ ) {
        int curLen = 0;
        int temp = 1;
        int j = i;

        while( 1 ) {
            temp = temp * j;
            if( n % temp == 0 ) {
                curLen++;
                if( curLen > maxLen ) {
                    maxLen = curLen;
                    head = j - curLen + 1;
                }
            }
            else {
                break;
            }
            j++;
        }
    }

    if( head == 0 ) {
        printf( "1\n%d", n );
        return 0;
    }
    printf( "%d\n", maxLen );
    for( int i = head; i < maxLen + head; i++ ) {
        if( i == head ) printf( "%d", i );
        else printf( "*%d", i );
    }
    return 0;
}