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>
#include <ext/pb_ds/assoc_container.hpp> // Common file
#include <ext/pb_ds/tree_policy.hpp> // Including tree_order_statistics_node_update
#define ff first
#define ss second
#define pb push_back
#define pf push_front
#define eb emplace_back
#define emp emplace
#define ins insert
#define mp make_pair
#define mt make_tuple
#define sz(s) (int)s.size()
#define forp(i,a,b) for( int i=a;i<=b;i++)
#define rep(i,n)    for( int i=0;i<n;i++)
#define ren(i,n)    for( int i=n-1;i>=0;i--)
#define forn(i,a,b) for( int i=a;i>=b;i--)
#define w(t) while(t)
#define on cout<<"\n"
#define os cout <<" "
#define o2(a,b) cout<<a<<" "<<b
#define o(a) cout << a
#define ppr(a) cout << a.first << " " << a.second
#define bitcount __builtin_popcount
#define gcd __gcd
#define all(v) v.begin(),v.end()
#define mem(n,m) memset(n,m,sizeof(n))
#define pii pair<int,int>
#define tiii tuple<int, int, int>
#define pll pair<long long,long long>
#define sii set<int>
#define sll set<long long>
#define vii vector<int>
#define vll vector<long long>
#define mii map<int,int>
#define mll map<long long, long long>
#define lob lower_bound
#define upb upper_bound
#define ret return
#define present(s,x) (s.find(x) != s.end())
#define cpresent(s,x) (find(all(s),x) != s.end())
#define ford(container, it) for(auto it = container.begin(); it != container.end(); it++)
#define fors(container, it, a, b) for(auto it = a; it != b; it++)
using namespace __gnu_pbds;
#define boost ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define MOD 1000000007
#define EPSILON 1e-9
#define PI acos(-1)
#define inf 1e17
#define int long long
#define siz 100005
#define SIZ 1000005
#define SIZE 200005
#define debug(a) cerr<<#a<<": ";for(auto i:a)cerr<<i<<" ";cerr<<'\n';
#define trace(a) cerr<<#a<<": "<<a<<"\n"
#define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>


typedef long long  ll;
typedef long double  ldo;
typedef double  db ;
using namespace std;
auto start = std::chrono::system_clock::now();
inline void skj()
{
  std::chrono::time_point<std::chrono::system_clock> end;
  end = std::chrono::system_clock::now();
  std::chrono::duration<double> elapsed_seconds = end - start;
  std::time_t end_time = std::chrono::system_clock::to_time_t(end);
  cerr<<"Time taken " << elapsed_seconds.count()*1000<<"\n";
}

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
//uniform_int_distribution<int> ud(1, 100); use this for random number generator

    //custom hash for unordered map
struct custom_hash {
    static uint64_t splitmix64(uint64_t x) {
        // http://xorshift.di.unimi.it/splitmix64.c
        x += 0x9e3779b97f4a7c15;
        x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
        x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
        return x ^ (x >> 31);
    }

    size_t operator()(uint64_t x) const {
        static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();
        return splitmix64(x + FIXED_RANDOM);
    }
};
inline int binexp(int a,int b,int m)
{
    int res=1;
    a%=m;
    while(b)
    {
        if(b&1)
            res=(res*1ll*a)%m;
        a=(a*1ll*a)%m;
        b>>=1;
    }
    return res;
}

inline void solve_me_senpai() {
    int r, D, x;
    cin >> r >> D >> x;
    rep(i, 10) {
        x = x * r - D;
        o(x);on;
    }
}
signed main()
{
    boost
    int t = 1;
    //cin >> t;
    //int a = 1;
    while(t--) {
        //cout << "Case #" << a  << ": ";
        solve_me_senpai();
        //a++;
    }
    //skj();
    return 0;
}