java-查找总和为给定数字的整数序列

Question

For n < 10. taking 4 for test. I need to print sequences of integers such that the sum of them is n.

class Test {
    public static void main(String args[]) {
        printAll(4);
    }

    public static void printAll(int k) {
        int count = 0;
        int a[] = new int[k];
        for (int i = 0; i < k; i++) {
            a[i] = 1;
        }
        int currentSum = 0;
        a[1] = 0;
        for (int i = 0; i < k; i++) {
            a[1]++;
            for (int j = 0; j < k - i; j++) {
                currentSum += a[i];
                count++;
                if (currentSum == k) {
                    for (int m = 0; m < count; m++) {
                        System.out.print(a[m] + " ");
                    }
                    System.out.println();
                }
            }
            count = 0;
        }
    }
}

for n = 4:

Desired output:

    1 1 1 1
    1 2 1 
    1 3

Output I get:

    1 1 1 1

This is actually the first step of code. Later i need to print

    2 1 1
    2 2

So basically my target is to get sum equal to 4(in this case) at each step of iteration

Answer 1

There is a pattern which I would use to solve this problem.

Notice this:

example: 4 //i starts at 0 and increments 1
1 1 1 1 = 1 1+i 1 1  //4-i numbers
1 2 1   = 1 1+i 1    //4-i numbers
1 3     = 1 1+i      //4-i numbers

example: 8
1 1 1 1 1 1 1 1 = 1 1+i 1 1 1 1 1 1 //8-i numbers
1 2 1 1 1 1 1   = 1 1+i 1 1 1 1 1   //8-i numbers
1 3 1 1 1 1     = 1 1+i 1 1 1 1     //8-i numbers
1 4 1 1 1       //and so on...
1 5 1 1
1 6 1
1 7

//Now we introduce a new level, with 'j' (j was 0 in previous example, now it is 1)
//We add i to j in the second position in the array
2 1 1 1 1 1 1 = 1+j j+i 1 1 1 1 //8-j-i numbers
2 2 1 1 1 1   = 1+j j+i 1 1 1   //8-j-i numbers
2 3 1 1 1     = 1+j j+i 1 1     //8-j-i numbers
2 4 1 1       = 1+j j+i 1       //8-j-i numbers
2 5 1
2 6

Putting this pattern together, we only use two different variables i and j, so we probably only need two for loops. (like you have.)

int exampleNumber = 8;
for (int j = 0; j < exampleNumber; j++)
{
    for (int i = 0; i < exampleNumber; i++)
    {

    }
}

Next, we know we start with an array of 8-j length, and then shrink it when i increments as well.

int exampleNumber = 8;
for (int j = 0; j < exampleNumber; j++)
{
    int jLength = exampleNumber - j; //NewLine
    for (int i = 0; i < exampleNumber; i++)
    {
        int iArray = new int[jLength - i]; //NewLine
    }
}

Now we just have to set the two important numbers. The one at the start and the one next to it.

int exampleNumber = 8;
for (int j = 0; j < exampleNumber; j++)
{
    int jLength = exampleNumber - j;
    for (int i = 0; i < exampleNumber; i++)
    {
        int iArray = new int[jLength - i];
        iArray[0] = 1 + j; //NewLine
        iArray[1] = j + i; //NewLine
    }
}

Finally, we just need to print what we have. We substitue 1's where the array is null.

int exampleNumber = 8;
for (int j = 0; j < exampleNumber; j++)
{
    int jLength = exampleNumber - j;
    for (int i = 0; i < exampleNumber; i++)
    {
        int iArray = new int[jLength - i];
        iArray[0] = 1 + j;
        iArray[1] = j + i;
        //printIntArray(iArray);
    }
}

public void printIntArray(int[] printArray)
{
    for (int i = 0; i < printArray.length(); i++)
    {
        int numberToPrint = printArray[i];
        if (numberToPrint == null || numberToPrint == 0)
            numberToPrint = 1;
        System.out.print(numberToPrint + " ");
    }
    System.out.println();
}

Answer 2

How about this less complicated solution:

public static void allSequencesSumToN(int n) {
    generate(n,"", n);
}

public static void generate(int no, String str, int orig) {
    if(no == 0) {
        System.out.println(str);
        return;
    }
    if(no < 0) {
        return;
    }
    for(int i = 1; i < orig ; i++) {
        waysUtil(no - i, str + i, orig);
    }

}

It might be slightly different than what you need, but you can adjust the result. Eg input 4, this will generate following output:

1111 112 121 13 211 22 31

All possible sequences, but with easy postprocessing you can achieve exactly what you need.