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;
typedef long long ll;
typedef vector<long long> vll;
typedef vector<pair<int,int>> vp;
typedef pair<int,int> pii;
#define INF (int)(1e9)
#define MAXX 1.1529215e+18
#define inf 999999
#define EPS (1e-7)
#define rep(i,n) for(int i=0; i<(int)(n);i++)
#define REP(i,n) for(int i=1;i<=(int)(n);i++)
#define FOR(i,k,n) for(int i=(k);i<(int)(n);i++)
#define ALL(a)  a.begin(),a.end()
#define RALL(a) a.begin(),a.end(),greater<int>()
#define ROT(a) a.begin(),a.begin()+1,a.end()
#define RROT(a) a.begin(),a.end()-1,a.end()
#define PB push_back
#define MP make_pair
#define PI acos(-1.0)
#define sz(a) a.size()
const ll MOD = 1e9+7;
const int MAX = 10000000;
template<typename T1, typename T2> inline bool chmin(T1 &a, T2 b) { if (a > b) { a = b; return 1; } return 0; }
template<typename T1, typename T2> inline bool chmax(T1 &a, T2 b) { if (a < b) { a = b; return 1; } return 0; }
template<typename T> T gcd(T a, T b) { if (b == 0) return a; return gcd(b, a % b); }
//cout << 'H' << endl;
//next_permutation(ALL(p));
//typedef priority_queue<int> pque;//これは大きい順
//typedef priority_queue<int, vector<int>, greater<int>> pque;//これは小さい順
//factorizeは、素因数分解用の関数

int main(){
	int r,d,x;
  cin >> r >> d >> x;
  rep(i,10){
    x = r*x - d;
    cout << x << endl;
  }

  return 0;
}