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 C++ (GCC 9.2.1) (judge verdict: Time Limit Exceeded)

#include<iostream>
#include<vector>
#include<string>
#include<algorithm>
#include<tuple>
#include<map>
#include<queue>
#include<math.h>
#include<numeric>
#include<iomanip>
#define rep(i, n) rep2(i,0,n)
#define rep2(i, s, n) for (int i = (s); i < (int)(n); i++)
#define PI acos(-1);
typedef long long ll;
using namespace std;
const int MOD = 1000000007;
//scanf("%d", &a); '\n'; cout << fixed << setprecision(10);

ll F(ll a, ll b, ll x) {
	return floor(a*x / b) - a * floor(x / b);
}

int main() {
	ios::sync_with_stdio(false);
	ll a, b, n;
	cin >> a >> b >> n;
	ll ans;
	if (F(a, b, 1) > F(a, b, n)) {
		ans = F(a, b, 1);
		rep2(i, 2, min(n + 1, ll(5e7))) {
			ans = max(ans, F(a, b, i));
		}
	}
	else {
		ans = F(a, b, n);
		for (ll i = n - 1; i > max(ll(1), ll(n - 5e7)); --i) {
			ans = max(ans, F(a, b, i));
		}
	}
	rep2(i,(1+n)/2 , min(n + 1, (1 + n) / 2+ll(6e7))) {
		ans = max(ans, F(a, b, i));
	}
	for (ll i = (1 + n) / 2; i > max(ll(1), (1 + n) / 2-ll(6e7)); --i) {
		ans = max(ans, F(a, b, i));
	}
	ans = max(ans, F(a, b, (1 + n) / 2));
	ans = max(ans, F(a, b, (1 + n) / 2+1));
	ans = max(ans, F(a, b, (1 + n) / 2-1));
	cout << ans;
	
}