srm 631 div 2 500

Author: keshavnandan

CatsOnTheLineDiv2.cpp

@@ -0,0 +1,70 @@
+#include <iostream>
+#include <sstream>
+#include <vector>
+#include <algorithm>
+#include <set>
+#include <map>
+#include <queue>
+#include <cstring>
+#include <climits>
+using namespace std;
+typedef pair<int,int> pi;
+typedef vector<int> vi;
+typedef vector<vi> vvi;
+typedef vector<string> vs;
+typedef vector<vs> vvs;
+
+class CatsOnTheLineDiv2 {
+        public:
+        string getAnswer(vector <int> p, vector <int> c, int t)
+        {
+            int n = p.size();
+            for(int c1 : c) if(2*t+1 < c1) return "Impossible";
+
+            vector<pi> v;
+            for(int i = 0; i < n; i++)
+                v.push_back(make_pair(p[i], c[i]));
+
+            sort(v.begin(), v.end());
+
+            for(int i = 0; i < n; i++){
+                p[i] = v[i].first;
+                c[i] = v[i].second;
+            }
+
+            int x = p[0]-t, y = x+c[0]-1;
+
+            for(int i = 1; i < n; i++){
+                int p1 = p[i], c1 = c[i];
+                x = max(y+1, p1-t);
+                y = x+c1-1;
+                if(x > p1+t || y > p1+t) return "Impossible";
+            }
+
+            return "Possible";
+        }
+
+// BEGIN CUT HERE
+	public:
+	void run_test(int Case) { if ((Case == -1) || (Case == 0)) test_case_0(); if ((Case == -1) || (Case == 1)) test_case_1(); if ((Case == -1) || (Case == 2)) test_case_2(); if ((Case == -1) || (Case == 3)) test_case_3(); if ((Case == -1) || (Case == 4)) test_case_4(); }
+	private:
+	template <typename T> string print_array(const vector<T> &V) { ostringstream os; os << "{ "; for (typename vector<T>::const_iterator iter = V.begin(); iter != V.end(); ++iter) os << '\"' << *iter << "\","; os << " }"; return os.str(); }
+	void verify_case(int Case, const string &Expected, const string &Received) { cerr << "Test Case #" << Case << "..."; if (Expected == Received) cerr << "PASSED" << endl; else { cerr << "FAILED" << endl; cerr << "\tExpected: \"" << Expected << '\"' << endl; cerr << "\tReceived: \"" << Received << '\"' << endl; } }
+	void test_case_0() { int Arr0[] = {0}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arr1[] = {7}; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); int Arg2 = 3; string Arg3 = "Possible"; verify_case(0, Arg3, getAnswer(Arg0, Arg1, Arg2)); }
+	void test_case_1() { int Arr0[] = {0}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arr1[] = {8}; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); int Arg2 = 2; string Arg3 = "Impossible"; verify_case(1, Arg3, getAnswer(Arg0, Arg1, Arg2)); }
+	void test_case_2() { int Arr0[] = {0, 1}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arr1[] = {3, 1}; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); int Arg2 = 0; string Arg3 = "Impossible"; verify_case(2, Arg3, getAnswer(Arg0, Arg1, Arg2)); }
+	void test_case_3() { int Arr0[] = {5, 0, 2}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arr1[] = {2, 3, 5}; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); int Arg2 = 2; string Arg3 = "Impossible"; verify_case(3, Arg3, getAnswer(Arg0, Arg1, Arg2)); }
+	void test_case_4() { int Arr0[] = {5, 1, -10, 7, 12, 2, 10, 20}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arr1[] = {3, 4, 2, 7, 1, 4, 3, 4}; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); int Arg2 = 6; string Arg3 = "Possible"; verify_case(4, Arg3, getAnswer(Arg0, Arg1, Arg2)); }
+
+// END CUT HERE
+
+ };
+
+// BEGIN CUT HERE
+int main(){
+
+        CatsOnTheLineDiv2 ___test;
+        ___test.run_test(-1);
+
+}
+// END CUT HERE

CatsOnTheLineDiv2.txt

@@ -0,0 +1,83 @@
+PROBLEM STATEMENT
+
+There are some cats sitting on a straight line that goes from the left to the right.
+You are given two vector <int>s position and count.
+For each valid i, there are count[i] cats initially sitting at the point position[i].
+
+
+
+
+During each minute, each cat chooses and performs one of three possible actions: it may stay in its place, move one unit to the left (i.e., from x to x-1), or move one unit to the right (i.e., from x to x+1).
+(Note that there are no restrictions. In particular, different cats that are currently at the same point may make different choices.)
+
+
+
+
+You are also given an int time.
+The goal is to rearrange the cats in such a way that each point contains at most one cat.
+Return "Possible" if it's possible to achive the goal in time minutes, and "Impossible" otherwise (quotes for clarity).
+
+
+DEFINITION
+Class:CatsOnTheLineDiv2
+Method:getAnswer
+Parameters:vector <int>, vector <int>, int
+Returns:string
+Method signature:string getAnswer(vector <int> position, vector <int> count, int time)
+
+
+CONSTRAINTS
+-position will contain between 1 and 50 elements, inclusive.
+-position and count will contain the same number of elements.
+-Each element of position will be between -1000 and 1000, inclusive.
+-All elements of position will be distinct.
+-Each element of count will be between 1 and 1000, inclusive.
+-time will be between 0 and 1000, inclusive.
+
+
+EXAMPLES
+
+0)
+{0}
+{7}
+3
+
+Returns: "Possible"
+
+There are 7 cats sitting at the origin in this case. There are also 7 different points that cats can reach in 3 minutes, so each cat can occupy a unique point. Thus, the answer is "Possible".
+
+1)
+{0}
+{8}
+2
+
+Returns: "Impossible"
+
+Unlike the first test case, in this case there are 8 cats for 7 available points. Thus, the answer is "Impossible".
+
+2)
+{0, 1}
+{3, 1}
+0
+
+Returns: "Impossible"
+
+
+
+3)
+{5, 0, 2}
+{2, 3, 5}
+2
+
+Returns: "Impossible"
+
+
+
+4)
+{5, 1, -10, 7, 12, 2, 10, 20}
+{3, 4, 2, 7, 1, 4, 3, 4}
+6
+
+Returns: "Possible"
+
+