a. easy b. greedy problem, seems as 330

contest-20231203/2951/main.cpp

@@ -0,0 +1,32 @@
+#include<stdcpp.h>
+using namespace std;
+class Solution {
+public:
+    vector<int> findPeaks(vector<int>& mountain) {
+        vector<int>rlt;
+        const int len = mountain.size();
+        for(int i = 1 ; i < mountain.size() - 1; i++){
+            if(mountain[i] > mountain[i-1] && mountain[i] > mountain[i+1])
+            rlt.push_back(i);
+        }
+
+        return rlt;
+    }
+};
+int main(){
+    Solution sol;
+
+    vector<int> ex1 = {2,4,4};
+    vector<int>rlt1 = sol.findPeaks(ex1);
+    for(int i = 0 ; i <rlt1.size();i++)
+        cout<<rlt1[i]<<' ';
+    cout<<endl;
+
+    vector<int> ex2 = {1,4,3,8,5};
+    vector<int>rlt2 = sol.findPeaks(ex2);
+    for(int i = 0 ; i <rlt2.size();i++)
+        cout<<rlt2[i]<<' ';
+    cout<<endl;
+
+    return 0;
+}

contest-20231203/2952/main.cpp

@@ -0,0 +1,42 @@
+#include<stdcpp.h>
+using namespace std;
+class Solution {
+public:
+    int minimumAddedCoins(vector<int>& coins, int target) {
+        const int len = coins.size(); 
+        sort(coins.begin(), coins.end());
+        int express = 0;
+        int idx = 0;
+        int ans = 0;
+        while(express < target){
+            if(idx < len && express+1 >= coins[idx]){
+                express += coins[idx++];
+            }else{
+                ans++;
+                express += express + 1;
+            }
+        }
+        cout<<ans<<endl;
+        return ans;
+    }
+};
+int main(){
+    Solution sol;
+
+    vector<int> ex1 = {1,4,10};
+    int target1 = 19;
+    sol.minimumAddedCoins(ex1,target1);
+
+    vector<int> ex2 = {1,4,10,5,7,19};
+    int target2 = 19;
+    sol.minimumAddedCoins(ex2,target2);
+
+    vector<int> ex3 = {1,1,1};
+    int target3 = 20;
+    sol.minimumAddedCoins(ex3,target3);
+
+    vector<int> ex4 = {1};
+    int target4 = 100000;
+    sol.minimumAddedCoins(ex4,target4);
+    return 0;
+}