Problem:
Score : 100 points

Problem Statement

You went shopping to buy cakes and donuts with X yen (the currency of Japan).

First, you bought one cake for A yen at a cake shop.
Then, you bought as many donuts as possible for B yen each, at a donut shop.

How much do you have left after shopping?

Constraints

1 \leq A, B \leq 1 000

A + B \leq X \leq 10 000

X, A and B are integers.

Input

Input is given from Standard Input in the following format:

X
A
B

Output

Print the amount you have left after shopping.

Sample Input 1

1234
150
100

Sample Output 1

84

You have 1234 - 150 = 1084 yen left after buying a cake.
With this amount, you can buy 10 donuts, after which you have 84 yen left.

Sample Input 2

1000
108
108

Sample Output 2

28

Sample Input 3

579
123
456

Sample Output 3

0

Sample Input 4

7477
549
593

Sample Output 4

405

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define IOS ios::sync_with_stdio(false); cin.tie(0);
#define FOR(i,s,n) for(int i = s; i < (n); i++)
#define REP(i,n) FOR(i,0,n)
#define ALL(n) (n).begin(), (n).end()
#define RALL(n) (n).rbegin(), (n).rend()
#define ATYN(n) cout << ( (n) ? "Yes":"No") << endl;
#define CFYN(n) cout << ( (n) ? "YES":"NO") << endl;
#define OUT(n) cout << (n) << endl;
using ll = long long;
using ull = unsigned long long;
using pii = pair<int,int>;

int main(void)
{
    IOS
    int x, a, b;
    cin >> x >> a >> b;
    x -= a;
    while(x - b >= 0) x -= b;
    OUT(x)

    return 0;
}