A1096. Consecutive Factors

最新推荐文章于 2021-02-25 14:55:25 发布

转载最新推荐文章于 2021-02-25 14:55:25 发布 · 128 阅读

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原文链接：http://www.cnblogs.com/zhuqiwei-blog/p/8520149.html

本文介绍了一种算法，用于找出正整数N的最大连续因子序列，通过遍历从2到sqrt(N)的所有可能因子，找到使得N能被整除的最长连续因子序列，并输出该序列。文章还提供了代码实现细节及注意事项。

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

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<algorithm>
 4 #include<math.h>
 5 using namespace std;
 6 int main(){
 7     int N, sqr, maxLen = -1, index;
 8     long long P;
 9     scanf("%d", &N);
10     sqr = (int)sqrt(N * 1.0);
11     for(int i = 2; i <= sqr + 1; i++){
12         int j = i;
13         P = 1;
14         while(1){
15             P *= j;
16             if(P < 0)
17                 break;
18             if(N % P != 0)
19                 break;
20             if(j - i + 1 > maxLen){
21                 maxLen = j - i + 1;
22                 index = i;
23             }
24             j++;
25         }
26     }
27     if(maxLen != -1){
28         printf("%d\n", maxLen);
29         printf("%d", index++);
30     }else{
31         printf("%d\n%d", 1, N);
32     }
33     for(int i = 1; i < maxLen; i++){
34         printf("*%d", index++);
35     }
36     cin >> N;
37     return 0;
38 }

View Code

总结：

1、测试630的序列，应该是测试 630 / 3， 630 /（3 * 4）， 630 / （3 * 4 * 5），而非测试630 / 3，630 / 4， 630 / 5。

2、考虑到无解的可能，比如素数13， index 与maxlen没有进入循环，应单独输出。

3、有可能sqrt(N)后有误差，应该判断到 sqrt(N + 1)而非 sqrt(N)。

4、只要有累乘，就应该用long long，并且考虑到溢出为负的可能。

转载于:https://www.cnblogs.com/zhuqiwei-blog/p/8520149.html