PAT (Advanced Level) Practise 1096 Consecutive Factors (20)-优快云博客

针对给定正整数N，本篇介绍如何找出N的最大连续因子数量及其最小连续因子序列。通过枚举初始因子并逐步验证的方法实现。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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:

Sample Output:

3
5*6*7

题意：给你一个数，问最长可以由几个连续的相邻数字相乘，长度相同时，选第一个数字最小的

解题思路：可以暴力枚举第一个数字，直到不能模除

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;

int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        int ma=0,ans=n,m=sqrt(n);;
        for(int i=2;i<=m;i++)
        {
            int k=i,kk=n;
            while(kk%k==0)
            {
                kk/=k;
                k++;
            }
            if(k-i>ma)
            {
                ma=k-i;
                ans=i;
            }
        }
        if(!ma)
        {
            printf("1\n%d\n",n);
            continue;
        }
        printf("%d\n",ma);
        printf("%d",ans);
        for(int i=ans+1;i<ans+ma;i++)
            printf("*%d",i);
        printf("\n");
    }
    return 0;
}