✨ adds digit dp to find numbers which sum is divisible by d

algorithms/dynamic-programming/digits/sum-digits-divisible-by-d.cpp

@@ -0,0 +1,31 @@
+const int MAXD(100), MAXKLEN(10000);
+string K;
+ll D;
+vi digits;
+ll _dp[MAXKLEN + 1][MAXD][2];
+const ll MOD(1'000'000'007);
+ll dp(int pos = 0, int md = 0, bool lim = 1) {
+  if (pos == len(digits)) return md == 0;
+
+  ll &ans = _dp[pos][md][lim];
+
+  if (ans != -1) {
+    return ans;
+  }
+
+  ans = 0;
+  int maxi = lim ? digits[pos] : 9;
+  for (int i = 0; i <= maxi; i++) {
+    int mdn = (md + i) % D;
+    ans = (ans + dp(pos + 1, mdn, lim & (i == maxi))) % MOD;
+  }
+
+  return ans;
+}
+
+ll solve(string x) {
+  digits.clear();
+  for (auto c : x) digits.emplace_back(c - '0');
+  memset(_dp, -1, sizeof _dp);
+  return dp();
+}

algorithms/dynamic-programming/digits/sum-digits-divisible-by-d.tex

@@ -0,0 +1,7 @@
+\subsubsection{Sum of digits divisible by $D$}
+
+Find the total of numbers $X$, $1 \leq X \leq K$ such that the sum of digits in $X$ is divisible by $D$, as $K$ can be very large read it as string !
+
+Solve $O(MAXKLEN \cdot MAXD)$.
+
+