Added chusky and hunger problem

Author: Rodolfo Quispe

chusky_hunger/problem_statement.txt

@@ -0,0 +1,41 @@
+Chusky and Hunger
+=========================
+
+Chusky lives in Dogland and as everyone in this city, he loves the local dish: Aguadito. There is a unique Polleria (a really nice place to eat Aguadito) in Dogland, hence, this eatery is really selective with its customers. To eat in the Polleria, Chusky must solve the next task: Given D boxes, each one has all integer numbers between L_i and R_i (1 <= i <= D) inclusive, indicate the K-th lowest unique existing element. Because there is big queue waiting after Chusky, he has just some seconds to give the correct answer, otherwise he will not be able to eat in the Polleria and will eat some boiled potatoes that has at home (he does not like it too much) .
+Help Chusky to eat an Aguadito.
+
+Input
+=====
+
+The first line of the input contains a number T (1 <= T <= 100) indicating the number of test cases. Each test case follows the next format:
+The first line of each test case contains a number D (1 <= D <= 100) representing the number of boxes Chusky receives, the second line contains a number K (1 <= K <= 10 ^ 15) representing the K- th number Chusky must guess, then D lines follows, each one in has two integers L_i and R_i (1 <= L_i <= R_i <= 10 ^ 15) representing that box i has all integers between L_i and R_i (1 <= i <= D) inclusive.
+
+Output
+=====
+
+For each test case, print the K-th lowest number between all boxes. Print -1 if no K-th lowest number exists.
+
+Example
+======
+
+Input
+
+2
+3
+2
+3 6
+3 4
+11 13
+1
+7
+1 5
+
+Output
+
+4
+-1
+
+Explanation:
+
+For the first test case, the first box has numbers {3, 4, 5, 6}, the second one has {3, 4} and the last one has {11, 13}, then all existing numbers are {3, 4 , 5, 6, 11, 13} and the 2nd lowest number is 4. In the second case the box has numbers {1, 2, 3, 4, 5}, thus there is no 7th lowest element.
+

chusky_hunger/solution.cpp

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+#include <bits/stdc++.h>
+//#define DEBUG
+using namespace std;
+typedef vector < int > vi; 
+typedef pair < long long int, long long int > pii;
+typedef vector < pii > vii;
+typedef long long int i64;
+int d;
+i64 k;
+bool byPairs(pii left, pii right)
+{
+    if(left.first != right.first)
+        return left.first < right.first;
+    return left.second < right.second;
+}
+i64 sol(vii &nums)
+{
+    sort(nums.begin(), nums.end(), byPairs);
+    
+    i64 ans = -1ll;
+    i64 last = -1ll;
+    int i = 0;
+    while(k>0ll && i < (int)nums.size())
+    {
+        if(last < nums[i].first) // no overlap
+        {
+            i64 len = nums[i].second - nums[i].first + 1ll;
+#ifdef DEBUG
+
+            cout<<nums[i].first<<" --- "<<nums[i].second<<" : "<<len<< "  > " <<k<<endl;
+#endif                  
+
+            if(len < k)
+                k-=len, last = nums[i].second, i++;
+            else
+            {
+                ans = nums[i].first + k - 1ll, k = 0ll; 
+
+            }
+        }
+        else // overlap
+        {
+            if(nums[i].second <=  last) i++;
+            else nums[i].first = last + 1;
+        }
+    }
+    return ans;
+}
+int main()
+{
+#ifdef DEBUG
+  clock_t begin = clock();
+#endif   
+  int test_cases;
+  scanf("%d", &test_cases);
+  while(test_cases--)
+  {
+    cin>>d>>k;
+    vii nums = vii();
+    for(int i = 0; i < d; i++)
+    {
+      pii x;
+      cin>>x.first>>x.second;
+      nums.push_back(x);
+    }
+    cout<< sol(nums) <<endl;
+  }
+#ifdef DEBUG
+  clock_t end = clock();
+  double elapsed_secs = double(end - begin) / CLOCKS_PER_SEC;
+  printf("%.10f\n", elapsed_secs);
+#endif   
+}
+
+