More than 2019 cattle off summer school camp (fifth) B generator 1

Portal

Meaning of the questions:

First input X0, X1, a, b and follow Xi = a * X (i-1) + b * X (i-2) (i> = 2)

After re-enter n, mod, so that you find a formula based on the value of X (n) to the modulo mod

 

answer:

I believe we will soon be able to quickly find the coefficient matrix of power

|X0  X1|  *  |0  a|   =  |X1 X2|

　　　　　|1  b|

It is to write fast power matrix according to the above formula

 

But we must pay attention to the number of bits n input on the topic has long been beyond the scope of a long long, so in order to find out the coefficient matrix of power is still a problem

We observed that the power of 2 ^ 121, which is a 1,2 * 10,1 * 100

2^121 = (2^1)*(2^10)*(2^2)*(2^10)*(2^1)

According to this feature, we can put a big power several times to unpack ^ _ ^

 

Code:

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef long long LL;
 4 const int size=1e6+5;
 5 char n[size];
 6 int x0,x1,a,b;
 7 typedef long long LL;
 8 LL mod;
 9 struct mat
10 {
11     int m[3][3];
12     mat(){for(int i=0;i<3;i++)for(int j=0;j<3;j++) m[i][j]=0;}
13     friend mat operator*(mat x,mat y)
14     {
15         mat ans;
16         for(int i=1;i<=2;i++)
17         {
18             for(int j=1;j<=2;j++)
19             {
20                 for(intk = 1 ; k <= 2 ; k ++ )
 21                  {
 22                      ans.m [i] [k] = (ans.m [i] [k] + 1LL * xm [i] [j] * ym [j] [ k])% change;
23                  }
 24              }
 25          }
 26          return year;
27      }
 28  };
29 inline quick_pow mat (mat is, int b)
 30  {
 31      matt years;
32      ans.m [ 1 ] [ 1 ] = 1 ; ans.m [ 2 ] [ 2 ] =1;
33     while(b)
34     {
35         if(b&1) ans=ans*a;
36         a=a*a;
37         b>>=1;
38     }
39     return ans;
40 }
41 int main()
42 {
43     scanf("%d%d%d%d",&x0,&x1,&a,&b);
44     scanf("%s%lld",n,&mod);
45     mat ori;
46     int len =strlen(n);
47     ori.m[1][2]=1,ori.m[2][1]=b;ori.m[2][2]=a;
48     mat ans;
49     ans.m[1][1]=1;ans.m[2][2]=1;
50     for(int i=len-1;i>=0;i--)
51     {
52         ans=ans*quick_pow(ori,n[i]-'0');
53         ori=quick_pow(ori,10);
54     }
55     printf("%lld\n",(1LL*x0*ans.m[1][1]+1LL*x1*ans.m[1][2])%mod);
56 }

View Code