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(void){
    long int a,b,n,lp,lq;
    long int max = 0;
    int maxNum;
    scanf("%ld%ld%ld",&a,&b,&n);
    if(0 < n < 100){
        for(long int i =0; i <=n; i++){
            lp = a*i/b;
            lq = i/b;
            long int ans = lp - a * lq;
            if(max < ans){
                max = ans;
            }
        }
    }else{
        long int data[20];
        data[0] = n/20;
        for(int i = 1; i < 10;i++){
            data[i] = data[i-1] + n/10;
        }
        for(int i = 0; i < 20;i++){
            lp = a*data[i]/b;
            lq = data[i]/b;
            long int ans = lp - a * lq;
            if(max < ans){
                maxNum = i;
                max = ans;
            }
        }
        for (int i = data[maxNum-1];i < data[maxNum + 1];i++){
            lp = a*i/b;
            lq = i/b;
            long int ans = lp - a * lq;
            if(max < ans){
                maxNum = i;
            }
        }
    }
    printf("%d",max);
    return 0;
}