POJ.3187 Backward Digit Sums

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

Idea:
Using binomial coeff

Source:

 int n, sum;  
 vector<int> a;  
 int c[10][10];  
 int main()  
 {  
      // calc binom coeff  
      c[0][0] = 1;  
      for (int i = 1; i < 10; i++) {  
           for (int j = 0; j < 10; j++) {  
                if (j == 0) c[i][j] = 1;  
                else c[i][j] = c[i - 1][j] + c[i - 1][j - 1];  
           }  
      }  
      // take input  
      cin >> n >> sum;  
      for (int i = 1; i <= n; i++) {  
           a.push_back(i);  
      }  
      //store ans  
      vector<string> ans;  
      do {  
           int d = 0;  
           for (int i = 0; i < n; i++) {  
                d += c[n - 1][i] * a[i];  
           }  
           if (d == sum) {  
                string str = "";  
                for (int i = 0; i < n; i++) str += (char)(a[i] + 'a' - 1);  
                ans.push_back(str);  
           }  
      } while (next_permutation(a.begin(), a.end()));  
      sort(ans.begin(), ans.end());  
      for (int i = 0; i < n; i++)  
      {  
           cout << (ans[0][i] - 'a' + 1) << " ";  
      }  
      cout << endl;  
      return 0;  
 }