[LintCode] k Sum

Given n distinct positive integers, integer k (k <= n) and a number target.

Find k numbers where sum is target. Calculate how many solutions there are?

Example

Given [1,2,3,4], k=2, target=5. There are 2 solutions:

[1,4] and [2,3], return 2.

http://www.lintcode.com/en/problem/k-sum/

dp[k][v] 表示使用了k个数和为v的方案数，那么对于新加入的一个数a，状态转移方程为：

dp[k][v] += dp[k-1][v-a];

下面是AC代码。

class Solution {
public:
    /**
     * @param A: an integer array.
     * @param k: a positive integer (k <= length(A))
     * @param target: a integer
     * @return an integer
     */
    
    int kSum(vector<int> A, int k, int target) {
        // wirte your code here
        vector<vector<int> > dp(k + 1, vector<int>(target + 1, 0));
        dp[0][0] = 1;
        for (auto a : A) {
            for (int kk = k; kk > 0; --kk) {
                for (int v = target; v >= a; --v) {
                    dp[kk][v] += dp[kk-1][v - a];
                }
            }
        }
        return dp[k][target];
    }
};