# POJ 3187 Backward Digits Sums

There are 1 to \(N\) digits (\(1\le N \le 10\)) in certain order. Add adjacent numbers together to get the next line, until the there is only one number left.(Just like Pascal's triangle) For example, there are 3 integers: $$ 1, 2, 3 $$ $$ 3, 5 $$ $$ 8 $$

So, given \(N\) and the final sum, find the lexicographically least ordering of integers.

Link: http://poj.org/problem?id=3187

### Notes:

Be aware of that the question is asking 1 to \(N\) digits, so we don't have to test all possible permutations from 1 to 10.

The results of

`next_permutation()`

in STL are in ascending order in default.

### Solution:

Use `next_permutation()`

to check all possibilities and stimulate the triangle additions. Since the permutation is already in lexicographical order, when we get the find the first result, it is the final answer.

```
#include<iostream>
#include<algorithm>
using namespace std;
int n, sum, ans[11], s[11];
void solve() {
for (int i = 1; i <= n; i++) ans[i-1] = i;
if (n == 1 && ans[0] == sum) {
cout << sum << endl;
return;
}
do {
for (int i = 0; i < n-1; i++)
s[i] = ans[i]+ans[i+1];
for (int i = n-2; i >= 0; i--) {
for (int j = 0; j < i; j++)
s[j] = s[j]+s[j+1];
}
if (s[0] == sum) {
for (int i = 0; i < n; i++) cout << ans[i] << " ";
cout << endl;
return;
}
} while (next_permutation(ans, ans+n));
}
int main(void) {
cin >> n >> sum;
solve();
return 0;
}
```