直接用昨天的完全背包推导公式，还是很简单的。

322. 零钱兑换

class Solution {
public:
    int coinChange(vector<int>& coins, int amount) {
        int n = coins.size();
        vector<int> dp(amount + 1, 1e9);
        dp[0] = 0;
        for (int i = 0; i < n; i ++)
        {
            for (int j = coins[i]; j <= amount; j ++)
            {
                dp[j] = min(dp[j], dp[j - coins[i]] + 1);
            }
        }
        return dp[amount] == 1e9 ? -1 : dp[amount];
    }
};

279. 完全平方数

class Solution {
public:
    // 先把是平方数的存到n
    int numSquares(int n) {
        vector<int> arrays;
        for (int i = 1; i * i <= n; i ++)
            arrays.push_back(i * i);

        vector<int> dp(n + 1, 1e9);
        dp[0] = 0;
        for (int i = 0; i < arrays.size(); i ++)
            for (int j = arrays[i]; j <= n; j ++)
                dp[j] = min(dp[j], dp[j - arrays[i]] + 1);

        return dp[n];
    }
};