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

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

input

3 4 2 1

output

input

3 4 3 1

output

#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 continue

typedef long long ll;

typedef unsigned long long ULL;

const int mod = ;

const double eps = 1e-;

const int inf = 0x3f3f3f3f;

const int N=*1e5+;

ll quick(ll a,ll n)

{

    ll res=;

    while(n){

        if(n&) res=(res*a)%mod;

        a=(a*a)%mod;

        n>>=;

    }

    return res;

}

ll yuan(ll n)

{

    return quick(n,mod-);

}

int main()

{

    ll a,b,n,x;

    while(~SC("%lld%lld%lld%lld",&a,&b,&n,&x)){

        ll ans=;

        ans+=quick(a,n)*x%mod;

        ans+=b*(quick(a,n)-)%mod*yuan(a-)%mod;

        PF("%lld\n",ans);

    }

    return ;

}

逆元求法：利用费马小定理

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

巴特西

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

最新文章

热门文章