catch up on various minor edits

Author: dcclyde

contests/2023-02-01_codeforces_848d2/C.cpp

@@ -890,12 +890,14 @@ void solve() {
     ckmin(K, seen.size());
 
     ll out = 0;
-    FOR(mask, 0, (1<<seen.size())) {
+    do {
+    // FOR(mask, 0, (1<<seen.size())) {
         uset<char> allowed;
-        FOR(b, 0, seen.size()) {
-            if ((mask>>b) & 1) {allowed.insert(seen[b]);}
-        }
-        if (allowed.size() != K) {continue;}
+        // FOR(b, 0, seen.size()) {
+        //     if ((mask>>b) & 1) {allowed.insert(seen[b]);}
+        // }
+        // if (allowed.size() != K) {continue;}
+        FOR(k, 0, K) {allowed.insert(seen[k]);}
 
         ll local = 0;
         ll curr = 0;
@@ -909,7 +911,8 @@ void solve() {
         }
         local += (curr * (curr+1)) / 2;
         ckmax(out, local);
-    }
+    // }
+    } while (next_combination(seen.begin(), seen.begin() + K, seen.end()));
 
     return ps(out);
 }

contests/2023-02-06_codeforces_841d2/E.thinking

@@ -40,4 +40,5 @@ I bet I need to use the problem structure some more.
 
 Can I do some kinda square root decomposition trick?
 
+
 .

contests/2023-02-09_codeforces_851d2/E.cpp

@@ -954,10 +954,10 @@ tcT> struct SegTree { // cmb(ID,b) = b
     }
 };
 template<class T>
-string to_string(SegTree<T> st) {
+string tsdbg(SegTree<T> st) {
     vector<T> out;
     FOR(k, st.n, st.n + st.orig_n) { out.push_back( st.seg[k] ); }
-    return to_string( out );
+    return tsdbg( out );
 }
 
 void solve() {
@@ -998,6 +998,7 @@ void solve() {
             FOR(j, 0, psz) {helper.push_back(st.seg[st.n+j]);}
             dbgR(helper);
         );
+        dbg(st);
         // st.detailed_printouts();
         running += dat[k];
         /**

ideas/WIP_ctrlf

@@ -8,9 +8,11 @@ You're about to be handed `corpus` of length N.
 You're guaranteed that `target` appears exactly once as a (contiguous) substring.
 You want to find that occurrence using the following imaginary tool (pretty similar to ctrl+F).
 
-To use the tool, you type some keystroke(s) matching [a-z ] or <deleteall> or <enter>. 
+To use the tool, you type some keystroke(s) matching [a-z ] or <deleteall> or <enter>.
 
 and your cursor automatically jumps to some random occurrence of that substring in the corpus. You can always add more characters later - like you could type 3 characters, decide you don't like what you see, and then add a 4th character.
 
 
+https://docs.google.com/document/d/1xp99xsVcjFmcCqXgqaeFuuehXykM6VzJPHAKL6wre8M/edit#
+
 .

templates/template.cpp

@@ -610,12 +610,12 @@ template<typename T>
 V<T> pqueue_to_iterable(priority_queue<T> PQ) { // PASS BY VALUE!
     V<T> working;
     while (!PQ.empty()) {working.push_back(PQ.top()); PQ.pop();}
-    reverse(all(working));
+    // reverse(all(working));
     return working;
 }
 
 template <typename T>
-string tsdbg(priority_queue<T>&& PQ) {
+string tsdbg(priority_queue<T> PQ) {
     return tsdbg(pqueue_to_iterable(PQ));
 }
 

todo

@@ -10,6 +10,13 @@ TODO:
 Rename tsdbg and tsish to repr and to_string in some order?
 
 
+Quick functions for max/min of a container? Similar to sumv?
+    (Does this already exist?)
+    (Yeah it definitely does... go remember how?)
+    Yeah, it's just *max_element(all(dat)).
+
+
+
 Maybe learn about jupyter server?
 
 Make a version of mi that uses lls?