Problem:
Score : 400 points

Problem Statement

Given are integers A, B, and N.

Find the maximum possible value of floor(Ax/B) - A × floor(x/B) for a non-negative integer x not greater than N.

Here floor(t) denotes the greatest integer not greater than the real number t.

Constraints

1 ≤ A ≤ 10^{6}

1 ≤ B ≤ 10^{12}

1 ≤ N ≤ 10^{12}

All values in input are integers.

Input

Input is given from Standard Input in the following format:

A B N

Output

Print the maximum possible value of floor(Ax/B) - A × floor(x/B) for a non-negative integer x not greater than N, as an integer.

Sample Input 1

5 7 4

Sample Output 1

2

When x=3, floor(Ax/B)-A×floor(x/B) = floor(15/7) - 5×floor(3/7) = 2. This is the maximum value possible.

Sample Input 2

11 10 9

Sample Output 2

9

Solution in PyPy3 (7.3.0) (judge verdict: Time Limit Exceeded)

# -*- coding: utf-8 -*-
import sys
import math
from bisect import bisect_left
from bisect import bisect_right
import collections
import copy
import heapq
from collections import defaultdict
from heapq import heappop, heappush
import itertools
input = sys.stdin.readline
from collections import defaultdict
from heapq import heappop, heappush

from decimal import *

##### リストの 二分木検索 #####
# bisect_left(lists, 3)
# bisect_right(lists, 3)

##### プライオリティキュー #####
# heapq.heapify(a) #リストaのheap化
# heapq.heappush(a,x) #heap化されたリストaに要素xを追加
# heapq.heappop(a) #heap化されたリストaから最小値を削除＆その最小値を出力

# heapq.heappush(a, -x) #最大値を取り出す時は、pushする時にマイナスにして入れよう
# heapq.heappop(a) * (-1) #取り出す時は、-1を掛けて取り出すこと

##### タプルリストのソート #####
# sorted(ans) #(a, b) -> 1st : aの昇順, 2nd : bの昇順
# sorted(SP, key=lambda x:(x[0],-x[1])) #(a, b) -> 1st : aの昇順, 2nd : bの降順
# sorted(SP, key=lambda x:(-x[0],x[1])) #(a, b) -> 1st : aの降順, 2nd : bの昇順
# sorted(SP, key=lambda x:(-x[0],-x[1])) #(a, b) -> 1st : aの降順, 2nd : bの降順

# sorted(SP, key=lambda x:(x[1])) #(a, b) -> 1st : bの昇順
# sorted(SP, key=lambda x:(-x[1])) #(a, b) -> 1st : bの降順

##### 累乗 #####
# pow(x, y, z) -> x**y % z

##### 割り算の切り上げ #####
# tmp = -(-4 // 3)

##### dict の for文 #####
# for k, v in d.items():
#     print(k, v)

def inputInt(): return int(input())
def inputMap(): return map(int, input().split())
def inputList(): return list(map(int, input().split()))
def inputStr(): return input()[:-1]

inf = float('inf')
mod = 1000000007

#-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-
#-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-

def main():
	A,B,N = inputMap()

	maxs = 0
	conts = 0
	for x in range(N,0,-1):
		tmp = math.floor(A * x / B) - A * math.floor(x / B)
		conts += 1
		if maxs < tmp:
			maxs = tmp
		if conts == 50000000:
			break

	print(maxs)



#-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-
#-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-
# nCr mod m
# rがn/2に近いと非常に重くなる
def combination(n, r, mod=10**9+7):
    r = min(r, n-r)
    res = 1
    for i in range(r):
        res = res * (n - i) * modinv(i+1, mod) % mod
    return res

# mを法とするaの乗法的逆元
def modinv(a, mod=10**9+7):
    return pow(a, mod-2, mod)

def egcd(a, b):
    if a == 0:
        return b, 0, 1
    else:
        g, y, x = egcd(b % a, a)
        return g, x - (b // a) * y, y

# nHr mod m
# 問題によって、combination()を切り替えること
def H(n, r, mod=10**9+7):
	# comb = Combination(n+r-1, mod)
	# return comb(n+r-1, r)
    return combination(n+r-1, r, mod)


class Combination:
    """
    O(n)の前計算を1回行うことで，O(1)でnCr mod mを求められる
    n_max = 10**6のとき前処理は約950ms (PyPyなら約340ms, 10**7で約1800ms)
    使用例：
    comb = Combination(1000000)
    print(comb(5, 3))  # 10
    """
    def __init__(self, n_max, mod=10**9+7):
        self.mod = mod
        self.modinv = self.make_modinv_list(n_max)
        self.fac, self.facinv = self.make_factorial_list(n_max)

    def __call__(self, n, r):
        return self.fac[n] * self.facinv[r] % self.mod * self.facinv[n-r] % self.mod

    def make_factorial_list(self, n):
        # 階乗のリストと階乗のmod逆元のリストを返す O(n)
        # self.make_modinv_list()が先に実行されている必要がある
        fac = [1]
        facinv = [1]
        for i in range(1, n+1):
            fac.append(fac[i-1] * i % self.mod)
            facinv.append(facinv[i-1] * self.modinv[i] % self.mod)
        return fac, facinv

    def make_modinv_list(self, n):
        # 0からnまでのmod逆元のリストを返す O(n)
        modinv = [0] * (n+1)
        modinv[1] = 1
        for i in range(2, n+1):
            modinv[i] = self.mod - self.mod//i * modinv[self.mod%i] % self.mod
        return modinv

#-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-
# dfs のサンプル
def dfs(graph,parent,counter,edge):
    stk = []
    stk.append(edge)
    while len(stk) > 0:
        p = stk.pop()
        for e in graph[p]:
            if parent[p] == e:
                continue
            else:
                parent[e] = p
                counter[e] += counter[p]
                stk.append(e)

if __name__ == "__main__":
	main()