矩阵快速幂
把指数按十进制拆开的快速幂。。。
#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
#define full(a, b) memset(a, b, sizeof a)
#define __fastIn ios::sync_with_stdio(false), cin.tie(0)
#define pb push_back
using namespace std;
typedef long long LL;
inline int lowbit(int x){ return x & (-x); }
inline int read(){
int ret = 0, w = 0; char ch = 0;
while(!isdigit(ch)){
w |= ch == '-', ch = getchar();
}
while(isdigit(ch)){
ret = (ret << 3) + (ret << 1) + (ch ^ 48);
ch = getchar();
}
return w ? -ret : ret;
}
inline int lcm(int a, int b){ return a / __gcd(a, b) * b; }
template <typename A, typename B, typename C>
inline A fpow(A x, B p, C lyd){
A ans = 1;
for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;
return ans;
}
LL x0, x1, a, b, p;
string n;
struct Matrix{
LL m[2][2];
Matrix(){
full(m, 0);
}
Matrix operator * (Matrix &a){
Matrix ret;
for(int i = 0; i < 2; i ++){
for(int j = 0; j < 2; j ++){
for(int k = 0; k < 2; k ++){
ret.m[i][j] = ret.m[i][j] + (m[i][k] * a.m[k][j] % p);
if(ret.m[i][j] >= p) ret.m[i][j] -= p;
}
}
}
return ret;
}
};
Matrix ori, t, e;
inline void init(){
e.m[0][0] = e.m[1][1] = 1;
ori.m[0][0] = x1, ori.m[1][0] = x0;
t.m[0][0] = a, t.m[0][1] = b, t.m[1][0] = 1;
}
inline Matrix fpow(Matrix &a, const string &n){
Matrix ret = e;
for(int i = n.size() - 1; i >= 0; i --){
int y = n[i] - '0';
while(y --) ret = ret * a;
Matrix tr = a * a;
a = tr * tr;
a = a * a * tr;
//for(int j = 1; j <= 9; j ++) a = a * tr;
}
return ret;
}
int main(){
__fastIn;
cin >> x0 >> x1 >> a >> b;
cin >> n >> p;
init();
t = fpow(t, n);
ori = t * ori;
cout << (ori.m[1][0] % p) << endl;
return 0;
}