Code this in java

Below is a clean, self-contained solution in Java that follows the rules you showed on-screen:

java
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class Main {

    private static final int MOD = 1_000_000_007;

    public static void main(String[] args) throws Exception {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

        /* ---------- input ---------- */
        int N = Integer.parseInt(br.readLine().trim());   // array size
        int L = Integer.parseInt(br.readLine().trim());   // minimum index gap
        int R = Integer.parseInt(br.readLine().trim());   // maximum index gap
        int X = Integer.parseInt(br.readLine().trim());   // required sum

        int[] A = new int[N];
        for (int i = 0; i < N; ++i) {
            A[i] = Integer.parseInt(br.readLine().trim());
        }

        /* ---------- DP ----------
           dp[i][s] = # of “interesting” **increasing** subsequences
                      that end at index i and have sum s.
           A subsequence can extend from an earlier j if:
                • A[j] ≤ A[i]          (non-decreasing values)
                • L ≤ i − j ≤ R        (index-gap constraint)
        --------------------------------*/
        long[][] dp = new long[N][X + 1];
        long answer = 0;

        for (int i = 0; i < N; ++i) {

            // single-element subsequence
            if (A[i] <= X) {
                dp[i][A[i]] = 1;
            }

            // extend from previous valid positions
            for (int j = Math.max(0, i - R); j <= i - L; ++j) {
                if (A[j] > A[i]) continue;           // must stay non-decreasing

                for (int s = 1; s + A[i] <= X; ++s) {
                    long ways = dp[j][s];
                    if (ways != 0) {
                        int newSum = s + A[i];
                        dp[i][newSum] = (dp[i][newSum] + ways) % MOD;
                    }
                }
            }

            answer = (answer + dp[i][X]) % MOD;      // collect all ending at i whose sum==X
        }

        System.out.println(answer);
    }
}

How it works

1. State definition dp[i][s] counts all valid subsequences finishing at i whose element-sum is s.
2. Transitions
   • Start a new subsequence at i if A[i] ≤ X.
   • For every earlier index j with the required distance (L ≤ i-j ≤ R) and non-decreasing value (A[j] ≤ A[i]), extend each subsequence ending at j by appending A[i].
3. Complexity
   • Time ≈ O(N · (R-L+1) · X) (≤ 500·500·1000 ≈ 2.5·10^8 worst-case, ok for the given limits).
   • Memory ≈ O(N·X) (500·1001 ≈ 0.5 M integers).

The algorithm respects every rule in the prompt:

• Increasing (A[j] ≤ A[i]).
• Index-gap (L ≤ distance ≤ R).
• Exact sum (== X).
• Modulo 10⁹+7 for large counts.

Compile (javac Main.java) and run (java Main) with input formatted exactly as shown in your screenshots.

java
static int get_ans(int N, int L, int R, int X, List<Integer> A) {
    final int MOD = 1_000_000_007;
    long[][] dp = new long[N][X + 1];
    long ans = 0;

    for (int i = 0; i < N; i++) {
        int ai = A.get(i);

        // start a new single-element subsequence
        if (ai <= X) dp[i][ai] = 1;

        // extend valid subsequences ending within [i-R, i-L]
        for (int j = Math.max(0, i - R); j <= i - L; j++) {
            if (A.get(j) > ai) continue; // maintain non-decreasing order
            for (int s = 1; s + ai <= X; s++) {
                long ways = dp[j][s];
                if (ways == 0) continue;
                dp[i][s + ai] = (dp[i][s + ai] + ways) % MOD;
            }
        }
        ans = (ans + dp[i][X]) % MOD; // count subsequences whose sum == X
    }
    return (int) ans;
}