问题
I'm making an math app for the android. In one of these fields the user can enter an int (no digits and above 0). The idea is to get all possible sums that make this int, without doubles (4+1 == 1+4 in this case). The only thing known is this one int.
For example:
Say the user enters 4, I would like the app to return:
- 4
- 3+1
- 2+2
- 2+1+1
- 1+1+1+1
Obviously 4 == 4 so that should be added too. Any suggestions as to how i should go about doing this?
回答1:
Here's a simple algorithm that purports to do that
from : http://introcs.cs.princeton.edu/java/23recursion/Partition.java.html
public class Partition { public static void partition(int n) { partition(n, n, ""); } public static void partition(int n, int max, String prefix) { if (n == 0) { StdOut.println(prefix); return; } for (int i = Math.min(max, n); i >= 1; i--) { partition(n-i, i, prefix + " " + i); } } public static void main(String[] args) { int N = Integer.parseInt(args[0]); partition(N); } }
回答2:
There are short and elegant recursive solution to generate them, but the following may be easier to use and implement in existing code:
import java.util.*;
public class SumIterator implements Iterator<List<Integer>>, Iterable<List<Integer>> {
// keeps track of all sums that have been generated already
private Set<List<Integer>> generated;
// holds all sums that haven't been returned by `next()`
private Stack<List<Integer>> sums;
public SumIterator(int n) {
// first a sanity check...
if(n < 1) {
throw new RuntimeException("'n' must be >= 1");
}
generated = new HashSet<List<Integer>>();
sums = new Stack<List<Integer>>();
// create and add the "last" sum of size `n`: [1, 1, 1, ... , 1]
List<Integer> last = new ArrayList<Integer>();
for(int i = 0; i < n; i++) {
last.add(1);
}
add(last);
// add the first sum of size 1: [n]
add(Arrays.asList(n));
}
private void add(List<Integer> sum) {
if(generated.add(sum)) {
// only push the sum on the stack if it hasn't been generated before
sums.push(sum);
}
}
@Override
public boolean hasNext() {
return !sums.isEmpty();
}
@Override
public Iterator<List<Integer>> iterator() {
return this;
}
@Override
public List<Integer> next() {
List<Integer> sum = sums.pop(); // get the next sum from the stack
for(int i = sum.size() - 1; i >= 0; i--) { // loop from right to left
int n = sum.get(i); // get the i-th number
if(n > 1) { // if the i-th number is more than 1
for(int j = n-1; j > n/2; j--) { // if the i-th number is 10, loop from 9 to 5
List<Integer> copy = new ArrayList<Integer>(sum); // create a copy of the current sum
copy.remove(i); // remove the i-th number
copy.add(i, j); // insert `j` where the i-th number was
copy.add(i + 1, n-j); // insert `n-j` next to `j`
add(copy); // add this new sum to the stack
} //
break; // stop looping any further
}
}
return sum;
}
@Override
public void remove() {
throw new UnsupportedOperationException();
}
}
You can use it like this:
int n = 10;
for(List<Integer> sum : new SumIterator(n)) {
System.out.println(n + " = " + sum);
}
which would print:
10 = [10] 10 = [6, 4] 10 = [6, 3, 1] 10 = [6, 2, 1, 1] 10 = [7, 3] 10 = [7, 2, 1] 10 = [8, 2] 10 = [9, 1] 10 = [5, 4, 1] 10 = [5, 3, 1, 1] 10 = [5, 2, 1, 1, 1] 10 = [8, 1, 1] 10 = [7, 1, 1, 1] 10 = [4, 3, 1, 1, 1] 10 = [4, 2, 1, 1, 1, 1] 10 = [6, 1, 1, 1, 1] 10 = [5, 1, 1, 1, 1, 1] 10 = [3, 2, 1, 1, 1, 1, 1] 10 = [4, 1, 1, 1, 1, 1, 1] 10 = [3, 1, 1, 1, 1, 1, 1, 1] 10 = [2, 1, 1, 1, 1, 1, 1, 1, 1] 10 = [1, 1, 1, 1, 1, 1, 1, 1, 1, 1]
回答3:
This is the mathematical concept known as partitions. In general, it's... difficult, but there are techniques for small numbers. A load of useful stuff linked from the wiki page.
回答4:
For a number N you know that the max number of terms is N. so, you will start by enumerating all those possibilities.
For each possible number of terms, there are a number of possibilities. The formula eludes me now, but basically, the idea is to start by (N+1-i + 1 + ... + 1) where i is the number of terms, and to move 1s from left to right, second case would be (N-i + 2 + ... + 1) until you cannot do another move without resulting in an unsorted combination.
(Also, why did you tagged this android again?)
回答5:
This is related to the subset sum problem algorithm.
N = {N*1, (N-1)+1, (N-2)+2, (N-3)+3 .., N-1 = {(N-1), ((N-1)-1)+2, ((N-1)-1)+3..}
etc.
So it's a recursive function involving substitution; whether that makes sense or not when dealing with large numbers, however, is something you'll have to decide for yourself.
回答6:
All of these solutions seem a little complex. This can be achieved by simply "incrementing" a list initialized to contain 1's=N.
If people don't mind converting from c++, the following algorithm produces the needed output.
bool next(vector<unsigned>& counts) {
if(counts.size() == 1)
return false;
//increment one before the back
++counts[counts.size() - 2];
//spread the back into all ones
if(counts.back() == 1)
counts.pop_back();
else {
//reset this to 1's
unsigned ones = counts.back() - 1;
counts.pop_back();
counts.resize(counts.size() + ones, 1);
}
return true;
}
void print_list(vector<unsigned>& list) {
cout << "[";
for(unsigned i = 0; i < list.size(); ++i) {
cout << list[i];
if(i < list.size() - 1)
cout << ", ";
}
cout << "]\n";
}
int main() {
unsigned N = 5;
vector<unsigned> counts(N, 1);
do {
print_list(counts);
} while(next(counts));
return 0;
}
for N=5 the algorithm gives the following
[1, 1, 1, 1, 1]
[1, 1, 1, 2]
[1, 1, 2, 1]
[1, 1, 3]
[1, 2, 1, 1]
[1, 2, 2]
[1, 3, 1]
[1, 4]
[2, 1, 1, 1]
[2, 1, 2]
[2, 2, 1]
[2, 3]
[3, 1, 1]
[3, 2]
[4, 1]
[5]
来源：https://stackoverflow.com/questions/7331093/getting-all-possible-sums-that-add-up-to-a-given-number