Problem:
Score : 200 points

Problem Statement

The development of algae in a pond is as follows.

Let the total weight of the algae at the beginning of the year i be x_i gram. For i≥2000, the following formula holds:

x_{i+1} = rx_i - D

You are given r, D and x_{2000}. Calculate x_{2001}, ..., x_{2010} and print them in order.

Constraints

2 ≤ r ≤ 5

1 ≤ D ≤ 100

D < x_{2000} ≤ 200

All values in input are integers.

Input

Input is given from Standard Input in the following format:

r D x_{2000}

Output

Print 10 lines. The i-th line (1 ≤ i ≤ 10) should contain x_{2000+i} as an integer.

Sample Input 1

2 10 20

Sample Output 1

30
50
90
170
330
650
1290
2570
5130
10250

For example, x_{2001} = rx_{2000} - D = 2 \times 20 - 10 = 30 and x_{2002} = rx_{2001} - D = 2 \times 30 - 10 = 50.

Sample Input 2

4 40 60

Sample Output 2

200
760
3000
11960
47800
191160
764600
3058360
12233400
48933560

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

#include <bits/stdc++.h>
using namespace std;

#define ll long long int
#define str string
#define sec .second

ll MOD = 1000000007,INFL = 1ll << 60,INF = 1 << 30;
ll A[10000000] = {};
ll X[1000000] = {};
ll aiti[10000];
int dy[]={0, 0, 1, -1},dx[]={1, -1, 0, 0};
vector <ll> v;
ll a = 0,b = 0,c = 0,n = 0,sum = 0, max = INF;
str S;
bool an = false,bn = false;

int findDigits(int n) {
        int i = 0;
        while (n > 0) {

                X[i] = n%10;
                n /= 10;
                i++;
        }
        return i;
}


int main(){
        cin.tie(0);
        ios::sync_with_stdio(false);
        //Code Line
cin >> a >> b >> c;
for(int i = 0;i < 10;i++){
  cout << a * c - b << endl;
  c =  a * c - b;
}
cin >> n;
  //Code Line
        return 0;
}