[AtCoder] F - Knapsack for All Subsets

Problem Statement

 

dp[i][j]: the number of subsets of A[0, i] whose sum is j.

dp[0][0] = 1, there is only 1 way of not picking anything from an empty array to achieve sum 0.

Answer is dp[N][S]

 

State Transition:

case 1, not using A[i] : all the subsets of A[0, i - 1] are also subsets of A[0, i], so dp[i][j] += dp[i - 1][j] * 2;

case 2, using A[i], dp[i][j] += dp[i - 1][j - A[i]]; 

 

 

    static void solve(int testCnt) {
        for (int testNumber = 0; testNumber < testCnt; testNumber++) {
            int n = in.nextInt(), s = in.nextInt(), mod = 998244353;
            int[] a = in.nextIntArrayPrimitiveOneIndexed(n);
            long[][] dp = new long[n + 1][s + 1];
            dp[0][0] = 1;
            for(int i = 1; i <= n; i++) {
                for(int j = 0; j <= s; j++) {
                    dp[i][j] = addWithMod(dp[i][j], dp[i - 1][j] * 2, mod);
                    if(j >= a[i]) {
                        dp[i][j] = addWithMod(dp[i][j], dp[i - 1][j - a[i]], mod);
                    }
                }
            }
            out.println(dp[n][s]);
        }
        out.close();
    }