Adding solution to J and S

J-Sushi.cpp

@@ -0,0 +1,44 @@
+#include<bits/stdc++.h>
+using namespace std;
+#define ll  long long int
+const int  M = 1e9 + 7;
+int n;
+double dp[301][301][301];
+
+
+// Concept used: Expected Values and Probability
+
+
+double f(int x, int y, int z)
+{
+    if(x < 0 || y < 0 || z < 0)
+        return 0;
+    if(x == 0 && y == 0 && z == 0)
+        return 0;
+    if(dp[x][y][z] > 0)
+        return dp[x][y][z];
+
+    double p0 = (n - (x + y + z))/(1.0 *n);
+    double p1 = x / (1.0 * n);
+    double p2 = y / (1.0 * n);
+    double p3 = z / (1.0 * n);
+
+    return dp[x][y][z] = (n + x*f(x-1, y, z) + y*f(x+1, y-1, z) + z*f(x, y+1, z-1))/(x+y+z);
+}
+
+int main()
+{
+
+    int i, m;
+
+    cin >> n;
+    int ones = 0, twos = 0, threes = 0;
+    for(int i = 0; i < n; i++)
+    {
+        cin >> m;
+        if(m == 1) ones++;
+        else if(m == 2) twos++;
+        else threes++;
+    }
+    cout << setprecision(10) << f(ones, twos, threes);
+}

S-Digit_Sum.cpp

@@ -0,0 +1,58 @@
+#include<bits/stdc++.h>
+using namespace std;
+#define int long long
+const int M = 1e9 + 7;
+const int N = (int)1e4;
+
+// Concept used: Digit DP
+
+int dp[N][2][100];
+
+
+int f(int ind, int is_tight, int sum_of_dig, string &k, int &d)
+{
+
+    sum_of_dig %= d;
+
+    if(ind >= k.size())
+        return ((sum_of_dig % d == 0) ? 1 : 0);
+
+    if(dp[ind][is_tight][sum_of_dig] != -1)
+        return dp[ind][is_tight][sum_of_dig];
+
+    int ans = 0;
+    int curr_dig = k[ind] - '0';
+
+    if(is_tight)
+    {
+        for(int i = 0; i <= curr_dig; i++)
+        {
+            if(i == curr_dig)
+                ans += f(ind + 1, 1, i + sum_of_dig, k, d);
+            else
+                ans += f(ind + 1, 0, i + sum_of_dig, k, d);
+            ans %= M;
+        }
+    }
+    else
+    {
+        for(int i = 0; i <= 9; i++)
+        {
+            ans += f(ind + 1, 0, i + sum_of_dig, k, d);
+            ans %= M;
+        }
+    }
+    return dp[ind][is_tight][sum_of_dig] = ans % M;
+}
+
+
+int32_t main()
+{
+    string k;
+    int d;
+    cin >> k >> d;
+    memset(dp, -1, sizeof(dp));
+
+    cout << (f(0, 1, 0, k, d) - 1 + M) % M;
+    // subtracting 1 because '0' would have been counted as an answer too
+}