Added Conton problem

Author: shakyaprashant

CONTON/QUESTION.txt

@@ -0,0 +1,26 @@
+One of the famous proofs of modern mathematics is Georg Cantor's demonstration that the set of rational numbers is enumerable. 
+The proof works by using an explicit enumeration of rational numbers as shown in the diagram below.
+1/1 1/2 1/3 1/4 1/5 ...
+2/1 2/2 2/3 2/4
+3/1 3/2 3/3
+4/1 4/2
+5/1
+In the above diagram, the first term is 1/1, the second term is 1/2, the third term is 2/1, the fourth term is 3/1, the fifth term is 2/2, and so on.
+Input
+The input starts with a line containing a single integer t <= 20, the number of test cases. t test cases follow.
+
+Then, it contains a single number per line.
+Output
+You are to write a program that will read a list of numbers in the range from 1 to 10^7 and will print for each number the corresponding term in Cantor's enumeration as given below.
+
+Example
+Input:
+3
+3
+14
+7
+
+Output:
+TERM 3 IS 2/1
+TERM 14 IS 2/4
+TERM 7 IS 1/4

CONTON/solution.c

@@ -0,0 +1,84 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+
+
+int main() 
+{
+	
+long long n,a,b,num,den;
+
+	int t;
+
+	cin>>t;
+
+	while(t--)
+
+	{
+	
+	cin>>a;
+
+	n=(-1+sqrt(1+(8*a)))/2;
+
+	b=(n*(n+1))/2;
+
+	if(a==b)
+
+	{
+
+	    if(a%2==0)
+
+	    {
+
+	        den=n;
+
+	        num=1;
+
+	        cout<<"TERM "<<a<<" IS "<<num<<"/"<<den<<endl;
+
+	    }
+
+	    else
+
+	    {
+
+	        den=1;
+
+	        num=n;
+
+	        cout<<"TERM "<<a<<" IS "<<num<<"/"<<den<<endl;
+
+	    }
+	}
+
+	else
+
+	{
+ 
+   	if(n%2==0)
+
+    	{
+
+    	  den=a-b;
+
+    	  num=n+2-den;
+
+    	  cout<<"TERM "<<a<<" IS "<<num<<"/"<<den<<endl;
+
+    	}
+
+    	else
+
+    	{
+
+    	    num=a-b;
+
+    	    den=n+2-num;
+
+    	    cout<<"TERM "<<a<<" IS "<<num<<"/"<<den<<endl;
+
+    	}
+	}
+	}
+	return 0;
+}