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

#define max  100000

//qsort(str, n, sizeof(int), up_comp);
int up_comp(const void *a, const void *b){return *(int*)a - *(int*)b;}
int down_comp(const void *a, const void *b){return *(int*)b - *(int*)a;}
int count_digit(int a){int i=0;while(a!=0){a=a/10;i++;}return i;}
int euclid( int a, int b ){int temp;if(a<b){temp=a;a=b;b=temp;}if(b<1)return -1;if(a%b==0) return b;return euclid(b,a%b);}
int int_sort( const void * a , const void * b ) {
  return strcmp(( char * )a , ( char * )b );
}

int main()
{
    int  f1, f2, ans, tmp=0;
    long int a, b, x, n;
    
    scanf("%ld%ld%ld", &a, &b, &n);
    
    for (x=0; x<=n; x++) {
        f1=floor((double)a*x/b);
        f2=a*floor((double)x/b);
        ans=f1-f2;
        if (ans>tmp) {
            tmp=ans;
        }
    }
    
    printf("%d", tmp);
    
    return 0;
}