2019牛客暑期多校训练营(第五场) generator 1

时间限制：C/C++ 2秒，其他语言4秒
空间限制：C/C++ 262144K，其他语言524288K
64bit IO Format: %lld

题目描述

   You are given four positive integers x0,x1,a,bx0​,x1​,a,b. And you know  for all i≥2.

   Given two positive integers n, and MOD, please calculate xnx_nxn​ modulo MOD.

   Does the problem look simple? Surprise! The value of n may have many many digits!

输入描述

The input contains two lines.
The first line contains four integers x0,x1,a,b(1≤x0,x1,a,b≤10^9).
The second line contains two integers n, MOD (1≤n<10(10^6),10^9<MOD≤2×10^9, n has no leading zero).

输出描述

Print one integer representing the answer.

输入

1 1 1 1 10 1000000001

输出

89

说明

The resulting sequence x is Fibonacci sequence. The 11-th item is 89.

输入

1315 521 20185 5452831 9999999999999999999999999999999999999 1000000007

输出

914730061

题意：已知x0,x1,a,b,n,mod,和公式，求对mod取余。

题解：十进制快速幂。              

代码：

#include<bits/stdc++.h>
using namespace std;
long long x0,x1,a,b,mod;
struct node
{
  long long Martix[2][2];
  node operator *(const node&n) const
  {
   node sum;
   int i,j,k;
   for(i=0;i<2;i++)
     for(j=0;j<2;j++)
     {
       sum.Martix[i][j]=0;
       for(k=0;k<2;k++)
         sum.Martix[i][j]=(sum.Martix[i][j]+Martix[i][k]*n.Martix[k][j])%mod;
     }
    return sum;
  }
};
node ans,t;
char n[1000005];
node quick(node A,int k);
int main()
{
  int i,k,l;
  scanf("%lld%lld%lld%lld",&x0,&x1,&a,&b);
  scanf("%s%lld",n, &mod);
  ans.Martix[0][0]=ans.Martix[1][1]=1;
  ans.Martix[1][0]=ans.Martix[0][1]=0;
  t.Martix[0][0]=a;t.Martix[0][1]=b;
  t.Martix[1][0]=1;t.Martix[1][1]=0;
  l=strlen(n);
  for(i=l-1;i>=0;i--)
  {
    ans=ans*quick(t,n[i]-'0');
    t=quick(t,10);
  }
  printf("%lld",(x1*ans.Martix[1][0]+x0*ans.Martix[1][1])%mod);
  system("pause");
  return 0;
}
node quick(node A,int k)
{
  node sum;
  sum.Martix[0][0]=sum.Martix[1][1]=1;
  sum.Martix[0][1]=sum.Martix[1][0]=0;
  while(k)
   {
     if(k&1) sum=sum*A;
     k>>=1;
     A=A*A;
   }
  return sum;
}