D. Iterated Linear Function

题目连接：

http://www.codeforces.com/contest/678/problem/D

Description

Consider a linear function f(x) = Ax + B. Let’s define g(0)(x) = x and g(n)(x) = f(g(n - 1)(x)) for n > 0. For the given integer values A, B, n and x find the value of g(n)(x) modulo 109 + 7.

Input

The only line contains four integers A, B, n and x (1 ≤ A, B, x ≤ 109, 1 ≤ n ≤ 1018) — the parameters from the problem statement.

Note that the given value n can be too large, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use long integer type.

Output

Print the only integer s — the value g(n)(x) modulo 109 + 7.

Sample Input

3 4 1 1

Sample Output

7

Hint

题意

g(0)(x) = x

g(n)(x)=f(g(n-1)(x))

f(x) = ax+b

给你a b n x

求g(n)(x)

题解：

推一下 发现 g(n)(x) = a^nx+a^(n-1)b+a^(n-2)b+…+ab+b

其实就是等比数列……

快速幂直接莽一波就好了

特判 a=1的时候

代码

#include<bits/stdc++.h>
using namespace std;
const long long mod = 1e9+7;
long long quickpow(long long  m,long long n,long long k)//返回m^n%k
{
    long long b = 1;
    while (n > 0)
    {
          if (n & 1)
             b = (b*m)%k;
          n = n >> 1 ;
          m = (m*m)%k;
    }
    return b;
}

int main()
{
    long long a,b,n,x;
    cin>>a>>b>>n>>x;
    if(a==1)
    {
        cout<<(n%mod*b+x)%mod<<endl;
        return 0;
    }
    long long ans=(quickpow(a,n,mod)-1+mod)%mod;
    ans=ans*quickpow(a-1,mod-2,mod)%mod*b%mod;
    ans=(ans+quickpow(a,n,mod)*x+mod)%mod;
    printf("%lld\n",ans%mod);
}