PAT1096:Consecutive Factors

Posted 2020-10-15 0kk470

1096. Consecutive Factors (20)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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
5*6*7

思路
 因为12！ < 2^31 < 13!。
 所以，因子大小不会超过12！,连续因子的长度不会超过12。
 从长度为12开始减小，令res为满足条件的最大因子，穷举所有可能直到满足N % res == 0就行。
 注：穷举的过程中res会溢出，不过不影响结果。。
代码

#include<iostream>
#include<math.h>
using namespace std;
int main()
{
    int N;
    while(cin >> N)
    {
        int maxnum = sqrt(N);
        for(int len = 12;len >= 1;len--)
        {
            for(int start = 2; start <= maxnum;start++)
            {
               int res = 1;
               for(int i = start;i - start < len;i++)
               {
                   res *= i;
               }
               if(N % res == 0)
               {
                   cout << len << endl << start;
                   for(int j = start + 1;j - start < len;j++)
                    cout << "*" << j;
                   return 0;
               }
            }
        }
        cout << 1 << endl;
        cout << N;
    }
}