poj3187 Backward Digit Sums

思路：

暴力乱搞。

实现：

#include <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;

int n, m, a[10], buf[10];

bool check()
{
    for (int i = 0; i < n; i++)
    {
        buf[i] = a[i];
    }
    for (int i = 0; i < n - 1; i++)
    {
        for (int j = 0; j < n - i - 1; j++)
            buf[j] += buf[j + 1];
    }
    return buf[0] == m;
}
int main()
{
    cin >> n >> m;
    for (int i = 1; i <= n; i++)
    {
        a[i - 1] = i;
    }
    do
    {
        if (check())
            break;
    } while (next_permutation(a, a + n));
    for (int i = 0; i < n; i++)
    {
        cout << a[i] << " ";
    }
    puts("");
    return 0;
}