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 <stdio.h>
#include <stdlib.h>
#include <time.h>
#include <string.h>
#include <unistd.h>
#include <math.h>
long long int ans=0;
//最大公約数
long long int getgcd(long long n, long long m)
{
  int r;
  if (n > m) {
    long long t = m;
    m = n;
    n = t;
  }
  while ((r = m % n) != 0) {
    m = n;
    n = r;
  }
return n;
}

//nの階乗を計算
long long factorial(long long n){
    long long int answer = 1;
while(n > 1){
  answer *= n;
  n--;
}
    return answer;
}

//組み合わせの数nCrを計算
long long calcNumOfCombination(long long  n, long long r){
    return factorial(n) / (factorial(r) * factorial(n-r));
}
 void yes(){
   printf("Yes\n");
 }

 void no(){
   printf("No\n");
 }

int main(int argc, char* argv[]){
	double a = 0, b = 0,c=0;
	scanf("%lf %lf %lf",&a,&b,&c);
  double max=0,e=0,m=0;
  for (double i = c; i!=c-100000000;i--) {
    if (i<0 || i>c) {
      break;
    }
    e=floor(a*i/b)-a*floor(i/b);
    if (e>max) {
      max=e;
      m=i;
    }
  }
  for (double i = 100000000; i!=1000000000;i++) {
    if (i<0 || i>c) {
      break;
    }
    e=floor(a*i/b)-a*floor(i/b);
    if (e>max) {
      max=e;
      m=i;
    }
  }
  printf("%.0lf\n",max);
  return 0;
}