Merge branch 'master' into maximum-building-2

Author: brebenelmihnea

content/3_Silver/Binary_Search.problems.json

@@ -174,8 +174,7 @@
       "isStarred": false,
       "tags": ["Binary Search"],
       "solutionMetadata": {
-        "kind": "autogen-label-from-site",
-        "site": "CF"
+        "kind": "internal"
       }
     },
     {

content/3_Silver/Prefix_Sums.mdx

@@ -185,7 +185,6 @@ These are also known as
 ## Solution - Static Range Sum
 
 <LanguageSection>
-
 <CPPSection>
 
 In C++ we can use
@@ -223,7 +222,6 @@ int main() {
 ```
 
 </CPPSection>
-
 <JavaSection>
 
 ```java
@@ -269,14 +267,13 @@ public class Main {
 ```
 
 </JavaSection>
-
 <PySection>
 
 ```py
 def psum(a):
 	psum = [0]
 	for i in a:
-		psum.append(psum[-1] + i)  # psum[-1] is the last element in the list
+		psum.append(psum[-1] + i)
 	return psum
 
 
@@ -289,8 +286,30 @@ for i in range(Q):
 	print(p[r] - p[l])
 ```
 
-</PySection>
+An alternative approach is to use [`itertools.accumulate`](https://docs.python.org/3/library/itertools.html#itertools.accumulate).
+Notice that we need to add a $0$ to the front of the array.
+If using a newer version, you can use the optional parameter `initial=0` as well.
+
+```py
+import itertools
+
+
+def psum(a):
+	return [0] + list(itertools.accumulate(a))
+
 
+# BeginCodeSnip{Same code as above}
+N, Q = map(int, input().split())
+a = list(map(int, input().split()))
+p = psum(a)
+
+for i in range(Q):
+	l, r = map(int, input().split())
+	print(p[r] - p[l])
+# EndCodeSnip
+```
+
+</PySection>
 </LanguageSection>
 
 ## Problems
@@ -299,8 +318,8 @@ for i in range(Q):
 
 <Warning>
 
-"Haybale Stacking" isn't submittable on the USACO website. Use
-[this link](https://www.spoj.com/problems/HAYBALE/) to submit.
+"Haybale Stacking" isn't submittable on the USACO website;
+Use [this link](https://www.spoj.com/problems/HAYBALE/) to submit instead.
 
 </Warning>
 

solutions/general/cf-4A.mdx

@@ -2,7 +2,7 @@
 id: cf-4A
 source: CF
 title: Watermelon
-author: Junwon Kim
+author: Junwon Kim, Ayush Shukla
 ---
 
 [Official Analysis (C++)](https://codeforces.com/blog/entry/95249)
@@ -34,4 +34,38 @@ int main() {
 ```
 
 </CPPSection>
+<JavaSection>
+
+```java
+import java.io.*;
+import java.util.*;
+
+public class Watermelon {
+	public static void main(String[] args) {
+		Kattio io = new Kattio();
+		int w = io.nextInt();
+		if (w % 2 == 0 && w > 2) {
+			io.println("YES");
+		} else {
+			io.println("NO");
+		}
+		io.close();
+	}
+
+	// CodeSnip{Kattio}
+}
+```
+
+</JavaSection>
+<PySection>
+
+```py
+w = int(input())
+if w % 2 == 0 and w > 2:
+	print("YES")
+else:
+	print("NO")
+```
+
+</PySection>
 </LanguageSection>

solutions/silver/cf-1520F1.mdx

@@ -0,0 +1,81 @@
+---
+id: cf-1520F1
+source: CF
+title: Guess the K-th Zero (Easy Version)
+author: Rameez Parwez
+---
+
+[Official Editorial (C++)](https://codeforces.com/blog/entry/90342)
+
+## Explanation
+
+We use binary search to find the location of the $k$th zero.
+
+To do this, we set a midpoint and then check if the number of zeros in the left half is greater than or equal to or strictly less than $k$.
+This allows us to narrow the location of the $k$th zero to either the left or right half.
+
+## Implementation
+
+**Time Complexity:** $\mathcal{O}(\log N)$
+
+<LanguageSection>
+<CPPSection>
+
+```cpp
+#include <bits/stdc++.h>
+using namespace std;
+
+int query(int r, int k) {
+	int res;
+	cout << "? " << 1 << " " << r << '\n';
+	cin >> res;
+	return r - res <= k;
+}
+
+int main() {
+	int n, t, k;
+	cin >> n >> t >> k;
+	k--;
+
+	int lo = 0;
+	int hi = n;
+	while (hi - lo > 1) {
+		int mid = (lo + hi) >> 1;
+		if (query(mid, k)) {
+			lo = mid;
+		} else {
+			hi = mid;
+		}
+	}
+
+	cout << "! " << hi << '\n';
+}
+```
+
+</CPPSection>
+<PySection>
+
+```py
+def query(r: int, k: int) -> bool:
+	print(f"? 1 {r}")
+	res = int(input())
+	return r - res <= k
+
+
+n, t = map(int, input().split())
+k = int(input()) - 1
+
+lo = 0
+hi = n
+while hi - lo > 1:
+	mid = (lo + hi) // 2
+	if query(mid, k):
+		lo = mid
+	else:
+		hi = mid
+
+print(f"! {hi}")
+```
+
+</PySection>
+</LanguageSection>