public int solve(int[] prices, int target) {
        /**
         * dp[i][j]：可选商品为前i个时，凑出总价为j的组合数
         * 求dp[n][target]
         */
        int n = prices.length;
        int[][] dp = new int[n+1][target+1];

        // 初始化，和为0的组合数为0
        for (int i = 0; i <= n; i++) dp[i][0] = 1;

        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= target; j++) {
                if (prices[i-1] <= j) {　　　// 第i个商品的价格小于等于当前总价，可以买/也可以不买
                    dp[i][j] = dp[i-1][j-prices[i-1]] + dp[i-1][j];
                } else {　　// 第i个商品的价格大于当前总价，不能买
                    dp[i][j] = dp[i-1][j];
                }
            }
        }
        return dp[n][target];
    }