Algorithm
Valera had two bags of potatoes, the first of these bags contains x (x ≥ 1) potatoes, and the second — y (y ≥ 1) potatoes. Valera — very scattered boy, so the first bag of potatoes (it contains x potatoes) Valera lost. Valera remembers that the total amount of potatoes (x + y) in the two bags, firstly, was not gerater than n, and, secondly, was divisible by k.
Help Valera to determine how many potatoes could be in the first bag. Print all such possible numbers in ascending order.
The first line of input contains three integers y, k, n (1 ≤ y, k, n ≤ 109; 
≤ 105).
Print the list of whitespace-separated integers — all possible values of x in ascending order. You should print each possible value of x exactly once.
If there are no such values of x print a single integer -1.
10 1 10
-1
10 6 40
2 8 14 20 26
Code Examples
#1 Code Example with C++ Programming
Code -
C++ Programming
#include <bits/stdc++.h>
using namespace std;
int x, k, n;
int main() {
scanf("%d %d %d", &x, &k, &n);
int f = (x / k) * k + k;
bool ok = false;
for(int i = f; i <= n; i += k) {
ok = true;
printf("%d ", i - x);
}
if(!ok)
puts("-1");
return 0;
}Input
Output
Demonstration
Codeforcess Solution Two Bags of Potatoes,A. Two Bags of Potatoes ,C,C++, Java, Js and Python ,Two Bags of Potatoes,Codeforcess Solution