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 <stdio.h>
#include <string.h>
#include <math.h>
int main()
{
    int a,b,n,flag=1;
    scanf("%d %d %d",&a,&b,&n);
    if(a*b>0) printf("%d",(a*n/b)-(a*(n/b)));
    if(a*b<0) 
    {
       for(int i=0;i<=n;i++)
       {
          if((a*i/b)-(a*(i/b))>0);
            flag=0;
       }
      if(flag==0)  printf("%d",(a*n/b)-(a*(n/b)));
      else printf("0");
    }
    //if(a*b<0&&n>b) printf("%d",(a*(n/b))-(a*n/b));
}