Print all unique combination of factors of a given number

后端 未结 9 719
挽巷
挽巷 2021-02-08 05:46

What is the most efficient algorithm to print all unique combinations of factors of a positive integer. For example if the given number is 24 then the output should be

相关标签:
9条回答
  • 2021-02-08 06:17
    bool isprime(int n){
    for(int i=2; i<=sqrt(n); i++)
        if(n%i==0)
            return false;
    return true;
    }
    
    void printprime(int n){
    
    int i,j,y=n;
    
    while(y%2==0){
        cout<<"2 * ";
        y/=2;
    }
    
    for(i=3; i<=sqrt(y); i+=2){
        while(y%i==0){
            cout<<i<<" * ";
            y/=i;
        }
    }
    
    if(y>2)
        cout<<y;
    }
    
    void allfacs(int n){
    
    int i;
    unordered_set<int> done;
    
    for(i=2; i<sqrt(n); i++){
        if(n%i==0){
            cout<<i<<" * "<<n/i<<endl;
    
            if(!isprime(i) && done.find(i) == done.end()){
                done.insert(i);
                printprime(i);
                cout<<n/i<<endl;
            }
            if(!isprime(n/i) && done.find(n/i) == done.end()){
                done.insert(n/i);
                cout<<i<< " * ";
                printprime(n/i);
                cout<<endl;
            }
        }
    }
    }
    
    0 讨论(0)
  • 2021-02-08 06:18

    Here is my solution based on @rici's ideas.

    def factors(number, max_factor=sys.maxint):
        result = []
    
        factor = min(number / 2, max_factor)
        while factor >= 2:
            if number % factor == 0:
                divisor = number / factor
    
                if divisor <= factor and divisor <= max_factor:
                    result.append([factor, divisor])
    
                result.extend([factor] + item for item in factors(divisor, factor))
    
            factor -= 1
    
        return result
    
    print factors(12) # -> [[6, 2], [4, 3], [3, 2, 2]]
    print factors(24) # -> [[12, 2], [8, 3], [6, 4], [6, 2, 2], [4, 3, 2], [3, 2, 2, 2]]
    print factors(157) # -> []
    
    0 讨论(0)
  • 2021-02-08 06:20
    #include<bits/stdc++.h>
    using namespace std;
    int n;
    // prod = current product of factors in combination vector
    // curr = current factor
    void fun(int curr, int prod, vector<int> combination )
    {
        if(prod==n)
        {
            for(int j=0; j<combination.size(); j++)
            {
                cout<<combination[j]<<" ";
            }
            cout<<endl; 
            return;
        }
    
        for(int i=curr; i<n; i++)
        {
            if(i*prod>n) break;
            if(n%i==0)
            {
                combination.push_back(i);
                fun(i, i*prod, combination);
                combination.resize(combination.size()-1);
            }
        }
    }
    
    int main()
    {
        cin>>n;
        vector<int> combination;
        fun(2, 1, combination);
        cout<<1<<" "<<n;
        return 0;
    }
    
    0 讨论(0)
  • 2021-02-08 06:37
    vector<unsigned int> GetAllFactors(unsigned int number)
    {
        vector<unsigned int> factors;
    
        for (int i = 2; i <= number; i++)
        {
            if (number % i == 0)
            {
                factors.push_back(i);
            }
        }
    
        return factors;
    }
    
    void DoCombinationWithRepetitionFactors(vector<unsigned int> allFactors, unsigned currentProduct, unsigned targetProduct, vector<unsigned int> currentFactorSet, unsigned currentFactorIndex)
    {
        if (currentProduct == targetProduct)
        {
            for (auto a : currentFactorSet)
            {
                cout << a << " , ";
            }
    
            cout << endl;
    
            return;
        }
    
    
        for (int i = currentFactorIndex; i < allFactors.size(); i++)
        {
            if (currentProduct * allFactors[i] <= targetProduct)
            {
                currentFactorSet.push_back(allFactors[i]);
                DoCombinationWithRepetitionFactors(allFactors, currentProduct * allFactors[i], targetProduct, currentFactorSet, i);
                currentFactorSet.pop_back();
            }
        }
    }
    
    0 讨论(0)
  • 2021-02-08 06:41

    I came up with this, seems easy to read and understand. Hope it helps!

    def getFactors(num):
    
        results = []
    
        if num == 1 or 0:
            return [num]
    
        for i in range(num/2, 1, -1):
    
            if (num % i == 0):
                divisor = num / i
    
                if(divisor <= i):
                    results.append([i, divisor])
    
                innerResults = getFactors(divisor)
    
                for innerResult in innerResults:
                    if innerResult[0] <= i:
                        results.append([i] + innerResult)
    
        return results
    
    0 讨论(0)
  • 2021-02-08 06:43

    Try this recursive approach that also takes in 2 more inputs namely a string to carry over the current value of i in for loop to perform subsequent reduction and also a temp int to know when not to print duplicate reversals i.e., 8*3 and 3*8.

    public static void printFactors(int number, String parentFactors, int parentVal) {
        int newVal = parentVal;
        for (int i = number - 1; i >= 2; i--) {
    
            if (number % i == 0) {
                if (newVal > i) {
                    newVal = i;
                }
                if (number / i <= parentVal && i <= parentVal
                        && number / i <= i) {
                    System.out.println(parentFactors + i + "*" + number / i);
                    newVal = number / i;
                }
    
                if (i <= parentVal) {
                    printFactors(number / i, parentFactors + i + "*", newVal);
                }
            }
    
        }
    
    }
    

    And call this method using printFactors(12,'',12)
    Let me know if you find flaws in this approach. Thanks!

    0 讨论(0)
提交回复
热议问题