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<stdlib.h>
#include<math.h>
#include<string.h>


int main(){

 long long int N,M,L;
  scanf("%lld %lld %lld",&N,&M,&L);

/*
  char A[];
  scanf("%s",A);
  */

  /*
  int A[N];
  int i;
  for(i=0;i<N;i++){
   scanf("%d",&A[i]);
   }
  */
long long int i;
long long int m,l;
m=0;
if (L<=100000){
  for (i=1;i<=L;i++){
   l=(long long int)(N*i/M)-N*(long long int)(i/M);
   if(m<=l){
    m=l;
   }
   else{
     printf("%lld",m);
     return(0);
   }
  }
     printf("%lld",m);
     return(0);
}

else{
  int j;
  j=0;
  long long int x;
  x = (long long int)(L/10);
 long long int b;
 while (b > 0){
  b = (long long int)(N*(x+1)/M)-N*(long long int)((x+1)/M)-(long long int)(N*i/M)-N*(long long int)(i/M);
  x = x + (long long int)(L/10);
  j+=1;
 }
 for (i=(long long int)(L/10)*(j-1);i<=(long long int)(L/10)*(j);i++){
   l=(long long int)(N*i/M)-N*(long long int)(i/M);
   if(m<=l){
    m=l;
   }
   else{
     printf("%lld",m);
     return(0);
   }
  }
     printf("%lld",m);
     return(0);
}












 return(0);
 }