POJ 3187 Backward Digit Sums 题解 《挑战程序设计竞赛》

 1 #include <iostream>
 2 #include <algorithm>
 3 
 4 using namespace std;
 5 
 6 int n;
 7 int sum;
 8 int mycase[11][11];
 9 
10 int main() {
11     cin >> n >> sum;
12     for (int i = 1; i <= n; i++) {
13         mycase[n][i] = i;
14     } 
15     do {
16         for (int i = n; i > 1; i--) {
17             for (int j = 1; j < i; j++) {
18                 mycase[i - 1][j] = mycase[i][j] + mycase[i][j + 1];
19             }
20         }
21         if (mycase[1][1] == sum) {
22             for (int i = 1; i <= n; i++) {
23                 cout << mycase[n][i] << " ";
24             }
25             break;
26         }
27     } while (next_permutation(mycase[n] + 1, mycase[n] + 1 + n));
28     return 0;
29 }