1096 Consecutive Factors (20分)

最新推荐文章于 2020-09-22 16:22:18 发布

zbchenchanghao

最新推荐文章于 2020-09-22 16:22:18 发布

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本文介绍了一个算法挑战，目标是找出给定正整数的最大连续因子序列，并列出最小的连续因子序列。通过遍历可能的起始因子并计算连续乘积，算法能够找到满足条件的连续因子，最终输出最长序列及其长度。

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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

题意：本题难点在于理解题意，意思为，给一个数因式分解，找到一个连续数最多的子式，若有多个连续数一样多的子式，则输出连续数最小的那个。一个例子包懂：例如 240 = 2 * 4 * 5 * 6 有三个连续数，240 = 2 * 3 * 4 * 5，有四个连续数，所以结果为2 * 3 * 4 * 5。

#include<iostream>

#include<vector>
#include<cmath>
using namespace std;

int main(){
    int n;
    cin >> n;  

    
    int sqrt_n = sqrt(n);
    int max_len = -1;
    int flag;
    for(int i = 2; i <= sqrt_n; i++){

        int sum = 1;
        
        for(int j = i; i <= sqrt_n; j++){
            sum = sum * j;
            if(n % sum == 0){
                if(max_len < j - i + 1){
                    max_len = j - i + 1;
                    flag = i;
                }

            }else
            {
                break;
            }
            

        }

    }
    if(max_len == -1){
        printf("1\n");
        printf("%d\n", n);
    }else
    {
        printf("%d\n", max_len);
        for(int i = flag; i < flag + max_len; i++){
            printf("%s%d", i == flag ? "" : "*", i);
        }
        printf("\n");
    }
    


    return 0;
}