1096. Consecutive Factors (20)-PAT甲级真题

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

题目大意：一个正整数N的因子中可能存在若干连续的数字。例如630可以分解为3*5*6*7，其中5、6、7就是3个连续的数字。给定任一正整数N，要求编写程序求出最长连续因子的个数，并输出最小的连续因子序列～

[Update v2.0] 由github用户littlesevenmo提供的更高效的解法：
不用算连续因子最多不会超过12个，也不需要三重循环，两重循环即可，直接去计算当前部分乘积能不能整除N
分析：1、如果只有一个因子，那么这个数只能为1或者质数。因此我们主要去计算两个及以上因数的情况。
2、在有两个及以上的数连乘中，因数的最大上限为sqrt(N) + 1
3、因此思路就是，不断构造连乘，看连乘的积是否是N的因数，如果是，则看这部分连乘的数的个数是否比已记录的多。
4、用变量first记录连乘的第一个数字，这里我把它赋初值为0，如果在寻找N的因数过程中，first没有改变，那么就表明N是1或者是一个质数～

#include <iostream>
#include <cmath>
using namespace std;
long int num, temp;
int main(){
    cin >> num;
    int first = 0, len = 0, maxn = sqrt(num) + 1;
    for (int i = 2; i <= maxn; i++) {
        int j; 
        temp = 1;
        for (j = i; j <= maxn; j++) {
            temp *= j;
            if (num % temp != 0) break;
        }
        if (j - i > len) {
            len = j - i;
            first = i;
        }
    }
    if (first == 0) {
        cout << 1 << endl << num;
    } else {
        cout << len << endl;
        for (int i = 0; i < len; i++) {
            cout << first + i;
            if (i != len - 1) cout << '*';
        }
    }
    return 0;
}

#include <iostream>

#include <cmath>

using namespace std;

long int num, temp;

int main(){

cin >> num;

int first = 0, len = 0, maxn = sqrt(num) + 1;

for (int i = 2; i <= maxn; i++) {

int j;

temp = 1;

for (j = i; j <= maxn; j++) {

temp *= j;

if (num % temp != 0) break;

}

if (j - i > len) {

len = j - i;

first = i;

}

if (first == 0) {

cout << 1 << endl << num;

} else {

cout << len << endl;

for (int i = 0; i < len; i++) {

cout << first + i;

if (i != len - 1) cout << '*';

}

return 0;

}

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