Educational Codeforces Round 13 D. Iterated Linear Function 逆元+公式+费马小定理_1

D. Iterated Linear Function

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Examples

input

3 4 1 1

output

7

input

3 4 2 1

output

25

input

3 4 3 1

output

79

 

 

#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <cmath>#include <vector>#include <queue>#include <map>#include <algorithm>#include <set>using namespace std;#define MM(a,b) memset(a,b,sizeof(a))#define SC scanf#define PF printf#define CT continuetypedef long long ll;typedef unsigned long long ULL;const int mod = 1000000007;const double eps = 1e-10;const int inf = 0x3f3f3f3f;const int N=2*1e5+10;ll quick(ll a,ll n){    ll res=1;    while(n){        if(n&1) res=(res*a)%mod;        a=(a*a)%mod;        n>>=1;    }    return res;}ll yuan(ll n){    return quick(n,mod-2);}int main(){    ll a,b,n,x;    while(~SC("%lld%lld%lld%lld",&a,&b,&n,&x)){        ll ans=0;        ans+=quick(a,n)*x%mod;        ans+=b*(quick(a,n)-1)%mod*yuan(a-1)%mod;        PF("%lld\n",ans);    }    return 0;}

逆元求法：利用费马小定理

http://blog.csdn.net/qq_21057881/article/details/51758437