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

#define mod 1000000007
#define limit 200001
typedef long long ll;
typedef long double ld;
#define rep(i,n) for(i = 0; i < n; i++)
#define max(a,b) {if(a>b){return a;}return b;}
ll min(a,b) {if(a<b){return a;}return b;}
int up_int(const void *a, const void *b){return *(int*)a - *(int*)b;}
int down_int(const void *a, const void *b){return *(int*)b - *(int*)a;}
int up_char(const void *left, const void *right) {return strcmp((char *)left,(char *)right);}

int main(){
    ll a,b,n,i;
    ll ans = 0;
    ll ans_tmp = 0;
    ll tmp = 0;
    ll max = 0;
    
    scanf("%lld %lld %lld",&a,&b,&n);
    ll x = min(b-1,n);
    
    ans += (a * x / b);
    tmp = x / b;
    ans -= a * tmp;
    printf("%lld\n",ans);
    
    return 0;
}