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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?
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。
1 class Solution { 2 public: 3 /** 4 * @param A: an integer array. 5 * @param k: a positive integer (k <= length(A)) 6 * @param target: a integer 7 * @return an integer 8 */ 9 10 int kSum(vector<int> A, int k, int target) { 11 // wirte your code here 12 vector<vector<int> > dp(k + 1, vector<int>(target + 1, 0)); 13 dp[0][0] = 1; 14 for (auto a : A) { 15 for (int kk = k; kk > 0; --kk) { 16 for (int v = target; v >= a; --v) { 17 dp[kk][v] += dp[kk-1][v - a]; 18 } 19 } 20 } 21 return dp[k][target]; 22 } 23 };
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原文地址：http://www.cnblogs.com/easonliu/p/4508185.html
