十进制快速幂（牛客多校第五场）-- generator 1

思路：

十进制快速幂。

 #include <stdio.h>//sprintf
 #include <cstdlib>////malloc  exit strcat itoa system("cls")
 #include <iostream>//pair
 #include <fstream>
 #include <bitset>
 //#include <map>   https://ac.nowcoder.com/acm/contest/885/B
 #include <vector>
 #include <stack>
 #include <set>
 #include <string.h>//strstr
 #include <string>
 #include <time.h>//srand(((unsigned)time(NULL))); Seed n=rand()%10 - 0~9;
 #include <cmath>
 #include <queue>//priority_queue<long long, vector<long long>, greater<long long> > q;
 #include <vector>
 //#include <math.h>
 //#include <windows.h>   https://www.nitacm.com/problem_show.php?pid=222
 #include <algorithm>//sort + unique : sz=unique(b+1,b+n+1)-(b+1);+nth_element(first, nth, last, compare)
 using namespace std;
 #define pr printf
 #define sc scanf
 #define fo(a,b,c) for((a)=(b);(a)<=(c);(a)++)//register long long i
 #define fr(a,b,c) for((a)=(b);(a)>=(c);(a)--)
 #define mem(a,b) memset((a),(b),sizeof((a)))
 const double e=2.718281828;
 const double PI=acos(-1.0);
 const double ESP=1e-;
 const long long inf=;
 const long long N=;
 long long mod;
 const long long MAXN=;

 long long x0,x1,a,b,MOD;
 struct MAT
 {
     long long mat[MAXN][MAXN];
     MAT()
     {
         mat[][]=mat[][]=;//对角矩阵 E;
         mat[][]=mat[][]=;
     }
     MAT operator*(const MAT &a)const
     {
         MAT b;
         long long i,j,k;
         mem(b.mat,);
         fo(i,,MAXN-)
         {
             fo(j,,MAXN-)
             {
                 fo(k,,MAXN-)
                 {
                     b.mat[i][j]=(b.mat[i][j]+mat[i][k]*a.mat[k][j]);
                     b.mat[i][j]%=mod;
                 }
             }
         }
         return b;
     }
 }transfer;

 MAT Mqpow(MAT x,long long n)
 {
     struct MAT temp;
     while(n)
     {
         if(n&)
             temp=temp*x;
         x=x*x;
         n>>=;
     }
     return temp;
 }

 char n[N];
 struct MAT mark[],start;
 int main()
 {
     sc("%lld%lld%lld%lld%s%lld",&x0,&x1,&a,&b,n,&MOD);
 //    mem(transfer.mat,0);
     mod=MOD;
     transfer.mat[][]=a,transfer.mat[][]=b;
     transfer.mat[][]=,transfer.mat[][]=;
     start.mat[][]=x1,start.mat[][]=x0;
     mark[]=transfer;
     for(int i=;i<=;++i)
         mark[i]=transfer*mark[i-];
     int l=strlen(n);
     struct MAT ans;
     for(int i=;i<=l-;++i)
     {
         ans=Mqpow(ans,);
         ans=ans*mark[n[i]-''];
     }
     start=ans*start;
     pr("%lld\n",start.mat[][]%MOD);
     return ;
 }