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If coin order matters, that is, each sequence is unique, the DP function is simple enough to make it 1D DP. But key is that order DOESN‘T matter, so we need to add one more state: ending coin. And for each DP advance step, we only put >= coins.
#include <iostream> #include <vector> #include <algorithm> #include <numeric> using namespace std; typedef unsigned long long ULL; int main() { unsigned n, m; cin >> n >> m; vector<ULL> v(m); for (unsigned i = 0; i < m; i++) cin >> v[i]; std::sort(v.begin(), v.end()); // dp[val][ending coin] vector<vector<ULL>> dp(n + 1, vector<ULL>(m, 0)); dp[0][0] = 1; for (unsigned i = 0; i <= n; i++) for (unsigned j = 0; j < m; j ++) { for (unsigned k = j; k < m; k ++) { if((ULL(i) + v[k]) <= ULL(n)) dp[ULL(i) + v[k]][k] += dp[i][j]; } } ULL ret = 0; for(unsigned i = 0; i < m; i ++) ret += dp[n][i]; cout << ret << endl; return 0; }
HackerRank - "The Coin Change Problem"
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原文地址：http://www.cnblogs.com/tonix/p/4609312.html
