PAT 1096. Consecutive Factors (20)

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:

35*6*7

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一开始想了半天，越想越复杂，后来瞄了一眼这道题的分数。。嗯。。既然只值20分，那就试试暴力破解吧！于是就用最简单的方法，从2开始连乘，一个个试过去，找到最长的。结果这样的代码就AC了。。。。代码如下：

#include <iostream>#include <algorithm>#include <cmath>#include <vector>using namespace std;int main(void){int N,max=0,temp,count;cin>>N;vector<int> result;vector<int> t;for(int i=2;i<=sqrt(N);i++){if(N%i!=0)continue;temp=N/i;count=1;t.clear();t.push_back(i);for(int j=i+1;j<=sqrt(N);j++){if(temp%j!=0)break;else{t.push_back(j);temp/=j;count++;}}if(count>max){max=count;result=t;}}if(result.size()==0){cout<<1<<endl<<N;return 0;}cout<<max<<endl;for(int i=0;i<result.size()-1;i++)cout<<result[i]<<"*";cout<<result[result.size()-1];}