# Backward Digit Sums (POJ 3187)

The basic idea is that we generate every possible permutations of length from 0 to 9 as the first line and quickly calculate the sum. Let's say so the items in the first line is . Then the value of the final sum at the bottom end of the triangle will be:

where means the number of combinations to pick objects from different objects.

import java.io.*;
import java.util.*;

public class BackwardDigitSums {
private static boolean nextPermuation(int[] nums) {
if (nums.length == 0) return false;
int i = nums.length - 1;
while (i > 0) {
if (nums[i - 1] < nums[i]) break;
i--;
}
if (i <= 0) return false;
int j = nums.length - 1;
while (nums[j] <= nums[i - 1] && j >= i) j--;
int t = nums[j];
nums[j] = nums[i - 1];
nums[i - 1] = t;
Arrays.sort(nums, i, nums.length);
return true;
}
private static int nextSubset(int comb) {
int x = comb & -comb, y = comb + x;
return ((comb & ~y) / x >> 1) | y;
}
private static int sumNumbers(int[] numbers) {
int N = numbers.length;
int sum = 0;
int t = 1;
for (int i = 0; i < N; i++) {
sum += t * numbers[i];
t = (N - i - 1) * t / (i + 1);
}
return sum;
}
private static void solve(int N, int sum) {
int comb = (1 << N) - 1;
int[] numbers = new int[N];
while (comb < (1 << 10)) {
int l = 0;
for (int i = 0; i < 10; i++) if (((comb >> i) & 1) == 1) numbers[l++] = i + 1;

do {
if (sum == sumNumbers(numbers)){
System.out.printf("%d", numbers[0]);
for (int i = 1; i < N; i++) {
System.out.printf(" %d", numbers[i]);
}
System.out.println();
return;
}
} while (nextPermuation(numbers));

comb = nextSubset(comb);
}
}
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int N = in.nextInt(), sum = in.nextInt();
solve(N, sum);
}
}