Add K-th Not Divisible By N

K-th_Not_Divisible_By_N/k-th_not_divisible_by_n.py

@@ -0,0 +1,71 @@
+"""
+Problem Statement:
+You are given two positive integers n and k. Print the k-th positive integer that is not divisible by n.
+
+Example:
+If n = 3 and k = 7, the numbers not divisible by 3 are: 1, 2, 4, 5, 7, 8, 10, 11, 13...
+The 7th number among them is 10.
+
+Input:
+The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases.
+Each test case consists of two positive integers n (2 ≤ n ≤ 10^9) and k (1 ≤ k ≤ 10^9).
+
+Output:
+For each test case, print the k-th positive integer that is not divisible by n.
+
+Examples:
+Input:
+6
+3 7
+4 12
+2 1000000000
+7 97
+1000000000 1000000000
+2 1
+
+Output:
+10
+15
+1999999999
+113
+1000000001
+1
+"""
+
+# Brute Force:
+
+t = int(input())  # number of test cases
+
+for _ in range(t):
+    n, k = list(map(int, input().split()))  # read n (divisor) and k (kth non-divisible number)
+
+    low = 1
+    high = n * k  # upper bound: the kth non-divisible number must be <= n*k
+    mid = 0
+
+    # binary search to find kth number not divisible by n
+    while low <= high: 
+        mid = (low + high) // 2  # middle of current search range
+
+        # count how many numbers from 1..mid are NOT divisible by n
+        f = mid - (mid // n)  
+
+        if f < k:
+            # fewer than k numbers found → move right
+            low = mid + 1  
+        elif f > k:
+            # more than k numbers found → move left
+            high = mid - 1
+        else:
+            # exactly k numbers found
+            # if mid itself is divisible by n, move one step left
+            # because we want a number not divisible by n
+            if mid % n == 0:
+                mid -= 1
+            break
+
+    print(mid)  # print the kth number not divisible by n
+
+
+
+