From 4131af8aba62c273ff28906b75404264bd443615 Mon Sep 17 00:00:00 2001
From: VidHongqiWu <vidwu@vidhongqiwu.com>
Date: Fri, 27 Oct 2023 21:44:22 -0700
Subject: [PATCH] Add a DP problem named New_Year_and_Hurry

---
 .DS_Store                                  | Bin 0 -> 10244 bytes
 Dynamic_Programming/New_Year_and_Hurry.cpp |  35 +++++++++++++++++++++
 2 files changed, 35 insertions(+)
 create mode 100644 .DS_Store
 create mode 100644 Dynamic_Programming/New_Year_and_Hurry.cpp

diff --git a/.DS_Store b/.DS_Store
new file mode 100644
index 0000000000000000000000000000000000000000..226ddeb58491a12ccdea858e0bcdb8fa06e89228
GIT binary patch
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diff --git a/Dynamic_Programming/New_Year_and_Hurry.cpp b/Dynamic_Programming/New_Year_and_Hurry.cpp
new file mode 100644
index 00000000..cf8b7031
--- /dev/null
+++ b/Dynamic_Programming/New_Year_and_Hurry.cpp
@@ -0,0 +1,35 @@
+/*Limak is going to participate in a contest on the last day of the 2016.
+The contest will start at 20:00 and will last four hours, exactly until midnight.
+There will be n problems, sorted by difficulty, i.e. problem 1 is the easiest and problem n is the hardest.
+Limak knows it will take him 5·i minutes to solve the i-th problem.
+Limak's friends organize a New Year's Eve party and Limak wants to be there at midnight or earlier.
+He needs k minutes to get there from his house, where he will participate in the contest first.
+How many problems can Limak solve if he wants to make it to the party?
+
+Input
+The only line of the input contains two integers n and k
+
+Output
+Print one integer, denoting the maximum possible number of problems Limak can solve so that he could get to the party at midnight or earlier.*/
+
+#include <iostream>
+#include <algorithm>
+#include <cstring>
+using namespace std;
+
+int main()
+{
+    int n, k;
+    cin >> n >> k;
+    int time_left = 240 - k;
+    int dp[time_left + 10];
+    memset(dp, 0, sizeof(dp));
+    for(int i = 1; i <= n; ++ i)
+        for(int j = time_left; j >= i * 5; -- j)
+        {
+           dp[j] = max(dp[j], dp[j - i * 5] + 1) ;
+        }
+    cout << dp[time_left] << endl;
+    return 0;
+}
+