Created 37.cpp

37.cpp

@@ -0,0 +1,65 @@
+#include<bits/stdc++.h>
+using namespace std;
+
+bool trap_the_traper(int ar[],int target)
+{
+    int a=ar[0],b=ar[1],c=ar[2];          // a,b,c store the button values 
+    ar[0]=target-a-b;                            //now the array contain all the new targets
+    ar[1]=target-a-c;
+    ar[2]=target-b-c;
+
+    int temp,temp2;
+    for(int i=0 ;i<3 ;i++)                  //loop to search for even permutation
+        for(int x=0 ;x<=target ;x++)
+            {
+            temp=a*x;
+            for(int y=0 ;b*y<=target-temp ;y++)
+                {
+                temp2=b*y;
+                for(int z=0 ;z*c<=target-temp-temp2 ;z++)   
+                    if(ar[i]==temp+temp2+(z*c) && (x+y+z)%2==0)
+                        return  false;                                            
+                }
+            }            
+
+    return true;
+}
+
+int main()
+{
+    int test_cases;
+    cin>>test_cases;
+    while(test_cases--)
+    {
+            int button_val[3];
+            for(int i=0 ;i<3 ;i++)
+                cin>>button_val[i];
+
+            int target;
+            cin>>target;
+
+        if(trap_the_traper(button_val,target)==true) 
+            cout<<"By going first one can force a win"<<endl;
+        else
+            cout<<"By going second one can force a win"<<endl;
+    }
+    return 0;
+}
+/*
+Consider the example of 21 floor with buttons 1,5,9.
+let us assume we are playing first,then If we anayalze the last step of the game we will realise that-
+the last turn shall be ours and last second turn shall be of hacker in the winning case,
+and thus the only possible ways of pressing buutons in last and last second turns are; 
+
+hacker  Us
+1       5   //1+5=6
+1       9   //1+9=10
+5       9   //5+9=14            (their reverse doesnt matter as end result will be same)
+
+Now,if we fix these last steps,then the total floors left for us to cover will be either 21-6 or 21-10 or 21-14
+i.e. 15,11,7.
+
+now if any permutations of (1*x)+(5*y)+(9*z) = 15 or 11 or 7 ,such that the sum of x+y+z=even.(it means even count of total button pressed)
+Then the hacker played first,and our assumtion of playing first was wrong but we still won.(even count means hacker played first)
+Also,if none of the above even permutation is found then by playing first we are sure to win.
+*/