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: Wrong Answer)

#include <bits/stdc++.h>
#define rep(i,n) for(int i=0;i<(int)(n);i++)
using namespace std;
using ll = long long ;
using P = pair<int,int> ;
using pll = pair<long long,long long>;
constexpr int INF = 1e9;
constexpr long long LINF = 1e17;
constexpr int MOD = 1000000007;
ll a,b;
ll n;

ll flr(ll x){
    return floor((double)(a*x)/b) - a*floor((double)(x)/b);
}

int main(){
    cin >> a >> b >> n;
    ll ans = 0;
    int cnt = 0;
    while(cnt < 10000000){
        ans = max(ans,flr(n-cnt));
        ++ cnt;
        if(n-cnt <= 0) break;
    }
    cnt = 0;
    for(ll i=(n/b)+b-1;i>0;i-=b){
        if(i <= n) ans = max(ans,flr(i));
        if(i <= 0) break;
        ++ cnt;
        if(cnt >= 10000000) break;
    }
    cout << ans << endl;
    return 0;
}