立志用最少的代碼做最高效的表達

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PAT甲級最優(yōu)題解——>傳送門

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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<2^31).

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
567

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分析：

• 既然某數(shù)能分解為幾個數(shù)字相乘，則這些數(shù)字必定是其因子。
• 在其因子中， 如果因子大于該數(shù)/2，則因子必定不連續(xù)，因此不需考慮。所以將范圍縮減為sqrt(n)
• 基于前兩點設計算法即可。

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#include<bits/stdc++.h> using namespace std; int main() {int n; cin >> n;int sqrt_n = sqrt(n);vector<int>m, tmp_v;for(int i = 2; i <= sqrt_n; i++) {int cp_i = i, sum = i;while(n % sum == 0 && cp_i <= sqrt_n) {tmp_v.push_back(cp_i);cp_i++;sum *= cp_i;}if(tmp_v.size() > m.size()) m = tmp_v;tmp_v.clear();}if(m.size() == 0) cout << 1 << '\n' << n;else {cout << m.size();for(int i = 0; i < m.size(); i++) cout << (i==0?"\n":"*") << m[i];}return 0; }

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