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 (GCC 5.4.1) (judge verdict: Runtime Error)

#include <stdio.h>
#include <stdlib.h>
#define inf 10e8 + 5

int main(void) {
  int n, m, i, j, k;
  scanf("%d%d", &n, &m);
  int l, r, d;
  int **wei;
  wei = (int**)malloc(sizeof(int*) * (n + 1));
  for(i = 0; i <= n; ++i) wei[i] = (int*)malloc(sizeof(int) * (n + 1));
  char f1[n + 1];
  for(i = 0; i <= n; ++i) for(j = 0; j <= n; ++j) wei[i][j] = inf;
  for(i = 0; i < m; ++i) {
    scanf("%d%d%d", &l, &r, &d);
    wei[l][r] = d;
  }
  for(i = 1; i <= n; ++i) {
    for(j = 1; j <= n; ++j) {
      for(k = 1; k <= n; ++k) {
        if(wei[i][k] != inf) {
          if(wei[i][j] != inf && wei[j][k] != inf) {
            if(wei[i][j] + wei[j][k] != wei[i][k]) { printf("No"); return 0;}
          }
        } else {
          if(wei[i][j] != inf && wei[j][k] != inf) wei[i][k] = wei[i][j] + wei[j][k];
        }
      }
    }
  }
  printf("Yes");
  for(i = 0; i <= n; ++i) free(wei[i]);
  free(wei);
  return 0;
}