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>
int main(){
  long long A,B,N,x;
  scanf("%lld %lld %lld",&A,&B,&N);
  if(N>B){
    x=B;
  }else{
    x=N;
  }
  int max=0;
  int num;
  if(x<1000000){
    while(x>0){
      num=(A*x)/B - A*(x/B);
    //printf("%lld\n",num );
      if(num>max){
        max=num;
      }
      x--;
    }
  }else{
    while(x-1000000>0){
      num=(A*x)/B - A*(x/B);
    //printf("%lld\n",num );
      if(num>max){
        max=num;
      }
      x--;
    }
  }
  printf("%d",max);
}