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 <math.h>

int main(){
	long max = 0, temp, A, B, N, x;
	int i;
	//long max = 0, temp;
	scanf("%ld %ld %ld", &A, &B, &N);

	if(B<N){
		for(i=B; i<=N; i++){
			//printf("%d\n",i);
			temp = (A*i)/B - A*(i/B);
			//printf("%ld\n",temp);
			if(max < temp) max = temp;
		}
	}
	else{
		for(i=1; i<=N; i++){
			//printf("%d\n",i);
			temp = (A*i)/B - A*(i/B);
			//printf("%ld\n",temp);
			if(max < temp) max = temp;
		}
	}

	printf("%ld", max);
	return 0;
}