Buggy python submission (verdict: Time Limit Exceeded):
import math,itertools,fractions,heapq,collections,bisect,sys,queue,copy

sys.setrecursionlimit(10**7)
inf=10**20
mod=10**9+7
dd=[(-1,0),(0,1),(1,0),(0,-1)]
ddn=[(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]

def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def LS(): return sys.stdin.readline().split()
def S(): return input()

def main():
  a,b,n=LI()

  # right=n
  # ans=math.floor(a*right/b)-a*math.floor(right/b)
  # left=0
  # while left<=right:
  #   half=(left+right)//2
  #   _ans=math.floor(a*half/b)-a*math.floor(half/b)
  #   if _ans==ans:
  #     return ans
  #   elif _ans>ans:
  #     right=half-1
  #     ans=_ans
  #   else:
  #     left=half+1

  ans=int(a*1/b)-a*int(1/b)
  x=int(math.sqrt(n))
  for i in range(max(0,n-x-100000000),n+1):
    ans=max(ans,int(a*i//b)-a*int(i//b))
  return ans

# main()
print(main())


Fixed submission:
import math,itertools,fractions,heapq,collections,bisect,sys,queue,copy

sys.setrecursionlimit(10**7)
inf=10**20
mod=10**9+7
dd=[(-1,0),(0,1),(1,0),(0,-1)]
ddn=[(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]

def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def LS(): return sys.stdin.readline().split()
def S(): return input()

def main():
  # floor(x/b)がギリギリ割り切れない辺りが一番大きくなりそう
  # -> 実験してみるとmath.floor(a*x/b)-a*math.floor(x/b)はループしてそう
  # -> 答えはfloor(x/b)がギリギリ割り切れないときの値に決め打つぜ！
  a,b,n=LI()
  x=min(n,b-1)
  return math.floor(a*x/b)-a*math.floor(x/b)

# main()
print(main())

Buggy python submission (verdict: Time Limit Exceeded):
import math,itertools,fractions,heapq,collections,bisect,sys,queue,copy

sys.setrecursionlimit(10**7)
inf=10**20
mod=10**9+7
dd=[(-1,0),(0,1),(1,0),(0,-1)]
ddn=[(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]

def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def LS(): return sys.stdin.readline().split()
def S(): return input()

def main():
  a,b,n=LI()

  # right=n
  # ans=math.floor(a*right/b)-a*math.floor(right/b)
  # left=0
  # while left<=right:
  #   half=(left+right)//2
  #   _ans=math.floor(a*half/b)-a*math.floor(half/b)
  #   if _ans==ans:
  #     return ans
  #   elif _ans>ans:
  #     right=half-1
  #     ans=_ans
  #   else:
  #     left=half+1

  ans=int(a*1/b)-a*int(1/b)
  for i in range(int(math.sqrt(n)),n+1):
    ans=max(ans,int(a*i/b)-a*int(i/b))
  return ans

# main()
print(main())


Fixed submission:
import math,itertools,fractions,heapq,collections,bisect,sys,queue,copy

sys.setrecursionlimit(10**7)
inf=10**20
mod=10**9+7
dd=[(-1,0),(0,1),(1,0),(0,-1)]
ddn=[(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]

def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def LS(): return sys.stdin.readline().split()
def S(): return input()

def main():
  # floor(x/b)がギリギリ割り切れない辺りが一番大きくなりそう
  # -> 実験してみるとmath.floor(a*x/b)-a*math.floor(x/b)はループしてそう
  # -> 答えはfloor(x/b)がギリギリ割り切れないときの値に決め打つぜ！
  a,b,n=LI()
  x=min(n,b-1)
  return math.floor(a*x/b)-a*math.floor(x/b)

# main()
print(main())

Buggy python submission (verdict: Time Limit Exceeded):
import math,itertools,fractions,heapq,collections,bisect,sys,queue,copy

sys.setrecursionlimit(10**7)
inf=10**20
mod=10**9+7
dd=[(-1,0),(0,1),(1,0),(0,-1)]
ddn=[(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]

def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def LS(): return sys.stdin.readline().split()
def S(): return input()

def main():
  a,b,n=LI()

  # right=n
  # ans=math.floor(a*right/b)-a*math.floor(right/b)
  # left=0
  # while left<=right:
  #   half=(left+right)//2
  #   _ans=math.floor(a*half/b)-a*math.floor(half/b)
  #   if _ans==ans:
  #     return ans
  #   elif _ans>ans:
  #     right=half-1
  #     ans=_ans
  #   else:
  #     left=half+1

  ans=math.floor(a*1/b)-a*math.floor(1/b)
  for i in range(int(math.sqrt(n)),n+1):
    ans=max(ans,math.floor(a*i/b)-a*math.floor(i/b))
  return ans

# main()
print(main())


Fixed submission:
import math,itertools,fractions,heapq,collections,bisect,sys,queue,copy

sys.setrecursionlimit(10**7)
inf=10**20
mod=10**9+7
dd=[(-1,0),(0,1),(1,0),(0,-1)]
ddn=[(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]

def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def LS(): return sys.stdin.readline().split()
def S(): return input()

def main():
  # floor(x/b)がギリギリ割り切れない辺りが一番大きくなりそう
  # -> 実験してみるとmath.floor(a*x/b)-a*math.floor(x/b)はループしてそう
  # -> 答えはfloor(x/b)がギリギリ割り切れないときの値に決め打つぜ！
  a,b,n=LI()
  x=min(n,b-1)
  return math.floor(a*x/b)-a*math.floor(x/b)

# main()
print(main())

Buggy python submission (verdict: Time Limit Exceeded):
import math,itertools,fractions,heapq,collections,bisect,sys,queue,copy

sys.setrecursionlimit(10**7)
inf=10**20
mod=10**9+7
dd=[(-1,0),(0,1),(1,0),(0,-1)]
ddn=[(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]

def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def LS(): return sys.stdin.readline().split()
def S(): return input()

def main():
  a,b,n=LI()

  # right=n
  # ans=math.floor(a*right/b)-a*math.floor(right/b)
  # left=0
  # while left<=right:
  #   half=(left+right)//2
  #   _ans=math.floor(a*half/b)-a*math.floor(half/b)
  #   if _ans==ans:
  #     return ans
  #   elif _ans>ans:
  #     right=half-1
  #     ans=_ans
  #   else:
  #     left=half+1

  ans=math.floor(a*1/b)-a*math.floor(1/b)
  for i in range(int(math.sqrt(n)),n+1):
    ans=max(ans,math.floor(a*i/b)-a*math.floor(i/b))
  return ans

# main()
print(main())


Fixed submission:
import math,itertools,fractions,heapq,collections,bisect,sys,queue,copy

sys.setrecursionlimit(10**7)
inf=10**20
mod=10**9+7
dd=[(-1,0),(0,1),(1,0),(0,-1)]
ddn=[(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]

def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def LS(): return sys.stdin.readline().split()
def S(): return input()

def main():
  # floor(x/b)がギリギリ割り切れない辺りが一番大きくなりそう
  # -> 実験してみるとmath.floor(a*x/b)-a*math.floor(x/b)はループしてそう
  # -> 答えはfloor(x/b)がギリギリ割り切れないときの値に決め打つぜ！
  a,b,n=LI()
  x=min(n,b-1)
  return math.floor(a*x/b)-a*math.floor(x/b)

# main()
print(main())