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.

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#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;
}